PLANE  AND  SPHERICAL 

TRIGONOMETRY 


BY 

GEORGE  N.  BAUER,  PH.D. 

N 

AND 

W.  E.  BROOKE,  C.E.,M.A. 

UNIVERSITY   OF   MINNESOTA 


SECOND   REVISED  EDITION 


D.   C.   HEATH  &  CO.,   PUBLISHERS 
BOSTON        NEW   YORK        CHICAGO 


PHYSIOS 


COPYRIGHT,  1907  AND  1917, 
BY  D.  C.   HEATH  &  Co. 


H7 


PREFACE 

THE  plan  and  scope  of  this  work  may  be  indicated  briefly 
by  the  following  characteristic  features  : 

Directed  lines  and  Cartesian  coordinates  are  introduced 
as  a  working  basis. 

Each  subject,  when  first  introduced,  is  treated  in  a  gen- 
eral manner  and  is  presented  as  fully  as  the  character  of 
the  work  demands.  This  avoids  the  necessity  of  treating 
the  same  subjects  several  times  for  the  purpose  of  modify- 
ing and  extending  certain  conceptions. 

Statements  in  the  form  of  theorems  and  problems  are 
used  freely  to  indicate  the  aim  of  various  articles  and  to 
define  the  data  clearly. 

Inverse  functions  are  treated  more  fully  than  is  customary. 

The  general  principles  governing  the  solution  of  triangles, 
the  solution  of  trigonometric  equations,  and  the  proof  of 
identities  are  carefully  presented. 

.  Special  attention  is  given  to  the  arrangement  of  compu- 
tations. 

The  sine  and  cosine  series  are  obtained  from  De  Moivre's 
Theorem,  thus  completing  the  line  of  development  which 
leads  to  the  calculation  of  the  trigonometric  functions. 

The  work  on  spherical  trigonometry  contains  the  devel- 
opment of  all  the  formulas  that  are  generally  used  in 
practical  astronomy. 

The  right  spherical  triangle  is  treated  from  two  points  of 
view :  as  a  special  case  of  the  oblique  triangle,  and  directly 

iii 


iv  PREFACE 

from  geometric  figures.     The   work  is   so  arranged  that 
either  view  may  be  presented  independently. 

The  solutions  of  the  oblique  spherical  triangle  by  means 
of  auxiliary  quantities,  characteristic  of  astronomy,  are 
included  as  interesting  mathematical  problems  and  as 
preparation  for  astronomical  work. 

PREFACE  TO  THE  SECOND  REVISED  EDITION 

The  second  edition  embodies  such  modifications,  rear- 
rangements, and  additions  as  have  been  suggested  by  experi- 
ence in  the  classroom. 

Due  to  numerous  requests  tables  have  been  added. 
Simple  three-place  tables  have  been  included.  By  their  use 
the  numerical  work  is  reduced  to  a  minimum,  thus  leaving 
the  student  free  to  give  more  attention  to  the  principles  in- 
volved in  the  solution  of  a  triangle.  Accordingly  a  few  tri- 
angles, suitable  for  three-place  table  or  slide  rule  computa- 
tion, have  been  introduced  at  the  beginning  of  the  list  of 
problems  under  the  right  triangle  and  also  under  the  oblique 
triangle. 

In  order  to  give  a  wider  range  to  computational  work  four- 
place  tables  as  well  as  five-place  tables  have  been  included. 

G.  N.  B. 

W.  E.  B. 

MINNEAPOLIS,  MINNESOTA, 
June,  1917. 


CONTENTS 

PLANE   TRIGONOMETRY 

CHAPTER  I 
RECTANGULAR  COORDINATES  AND  ANGLES 

ABT.  PAOB 

1.  Introduction 1 

RECTANGULAR  COORDINATES 

2.  Directed  lines .1 

3.  Lines  of  reference 2 

4.  Quadrants 2 

5.  Coordinates  of  a  point 2 

6.  Signs  of  coordinates        .        . 3 

7.  Exercises 3 

ANGLES 

8.  Magnitude  of  angles 4 

9.  Direction  of  rotation.     Positive  and  negative  angles      .        .  4 

10.  Initial  and  terminal  lines 6 

11.  Sign  of  terminal  line 5 

12.  Algebraic  sum  of  two  angles 6 

13.  Measurement  of  angles 6 

14.  Circular  or  radian  measure 7 

15.  Value  of  radian 7 

16.  Relation  between  degree  and  radian 7 

17.  Relation  between  angle,  radius,  and  arc         .        .        .        .9 

18.  Linear  and  angular  velocity 9 

19.  Examples 10 

v 


CONTENTS 


CHAPTER   II 
TRIGONOMETRIC  FUNCTIONS 

ART.  PAGE 

20.  Trigonometric  functions  introduced 12 

21.  Definitions  of  the  trigonometric  functions      ....  12 

22.  Values  of  the  trigonometric  functions  of  30°,  45°,  and  60°    .  13 

23.  Values  of  the  trigonometric  functions  of  120°,  135°,  and  150°  15 

24.  Signs  of  the  trigonometric  ratios 17 

25.  Trigonometric  functions  are  single  valued      ....  18 

26.  A  given  value  of  a  trigonometric  function  determines  an 

infinite  number  of  angles 19 

27.  Examples        . 20 


CHAPTER   III 

RIGHT  TRIANGLES 

28.  Statement  of  problem 22 

29.  Application  of  the  definitions  of  the  trigonometric  functions 

to  the  right  triangle 22 

30.  Trigonometric  tables       .        . 23 

31.  Formulas  used  in  the  solution  of  right  triangles     ...  24 

32.  Selection  of  formulas 25 

33.  Check  formulas 25 

34.  Suggestions  on  solving  a  triangle 25 

35.  Illustrative  examples 26 

36.  Examples         ..........  29 

37 .  Oblique  triangles 30 

38.  Applications 31 


CHAPTER,  IV 

VARIATIONS  OF  THE  TRIGONOMETRIC  FUNCTIONS 
REDUCTION  OF   FUNCTIONS   OF  n90°±« 

39.  Values  of  functions  at  0°,  90°,  180°,  270°,  and  360°       .        .  34 

40.  Variation  of  the  functions 35 

41.  Graphical  representation 39 

42.  Periodicity  of  the  trigonometric  functions      ....  40 

43.  Examples 41 


CONTENTS  vii 

ABT  PAGE 

44.  Use  of  formulas 42 

45.  Functions  of  —  a  in  terms  of  functions  of  a  .        .        .        .42 

46.  Functions  of  90°  -f  a  in  terms  of  functions  of  a  .        .43 

47.  Functions  of  90°  —  a  in  terms  of  functions  of  a  .        .44 

48.  Functions  of  180°  —  a  in  terms  of  functions  of  a    .        .        .45 

49.  Laws  of  reduction .        .        .46 

50.  Examples 46 

CHAPTER  V 
FUNDAMENTAL  KELATIONS.     LINE   VALUES 

51.  General  statement 47 

FUNDAMENTAL  RELATIONS 

52.  Development  of  formulas 47 

53.  The  use  of  exponents 48 

54.  Trigonometric  identities 49 

55.  Trigonometric  equations 49 

56.  Examples .51 

LINE  VALUES 

57.  Representation  of  the  trigonometric  functions  by  lines          .      53 

58.  Variations  of  the  trigonometric  functions  as  shown  by  line 

values 55 

59.  Fundamental  relations  by  line  values 66 

60.  Examples 57 

CHAPTER  VI 

FUNCTIONS   OF  THE   SUM   OF  TWO   ANGLES 
DOUBLE  ANGLES.     HALF   ANGLES 

61.  Statement  of  problem 61 

62.  The  sine  of  the  sum  of  two  acute  angles  expressed  in  terms 

of  the  sines  and  cosines  of  the  angles       ....  61 

63.  The  cosine  of  the  sum  of  two  acute  angles      ....  62 

64.  Importance  of  formulas  ........  62 

65.  Generalization  of  formulas 63 

66.  Tangent  of  the  sum  of  two  angles 65 

67.  Cotangent  of  the  sum  of  two  angles 65 

68.  Addition  formulas  .  65 


viii  CONTENTS 

ART.  PAGB 

69.  Sine,  cosine,  tangent,  and  cotangent  of  the  difference  of  two 

angles 66 

70.  Exercises .66 

71.  Double  angles 67 

72.  Half  angles 68 

73.  Sum  and  difference  of  two  sines  and  of  two  cosines       .        .      69 

74.  Equations  and  identities 70 

75.  Examples .        .71 

CHAPTER   VII 
INVERSE  FUNCTIONS 

76.  Statement  of  problem 74 

77.  Fundamental  idea  of  an  inverse  function       ....      74 

78.  Multiple  values  of  an  inverse  function    .         .        .  75 

79.  Principal  values .77 

80.  Interpretation  of  sin  sin-1  a  and  sin-1  sin  a     .        .        .        .77 

81.  Application  of  the  fundamental  relations  to  angles  expressed 

as  inverse  functions 78 

82.  The  value  of  any  function  of  an  inverse  function  ...      79 

83.  Some  inverse  functions  expressed  in  terms  of  other  inverse 

functions .81 

84.  Relations  between  inverse  functions  derived  from  the  formulas 

for  double  angles,  half  angles,  and  the  addition  formulas      82 
86.  Examples 82 

CHAPTER   VIII 
OBLIQUE   TRIANGLE 

86.  General  statement 84 

87.  Law  of  sines 84 

88.  Law  of  tangents 86 

89.  Cyclic  interchange  of  letters 85 

90.  Law  of  cosines 85 

91.  Sine  of  half  angle  in  terms  of  sides  of  triangle       ...  86 

92.  Cosine  of  half  angle  in  terms  of  sides  of  triangle  ...  87 

93.  Tangent  of  half  angle  in  terms  of  sides 88 

94.  Area  of  plane  triangle  in  terms  of  two  sides  and  included 

angle 89 

95.  Area  of  triangle  in  terms  of  a  side  and  two  adjacent  angles  .  89 


CONTENTS 


96.  Area  in  terms  of  sides 90 

97.  Formulas  for  solving  an  oblique  triangle      ....  90 

98.  Check  formulas 91 

99.  Illustrative  problems 92 

100.  The  ambiguous  case 95 

101.  Examples 98 

102.  MISCELLANEOUS  EXERCISES 102 

CHAPTER   IX 

DE  MOIVRE'S  THEOREM  WITH  APPLICATIONS 

103.  Introduction  to  chapter 113 

104.  Geometric  representation  of  a  complex  number  .        .        .  113 

105.  De  Moivre's  theorem 114 

106.  Geometric  interpretation 115 

107.  Applications  of  De  Moivre's  theorem 116 

108.  Cube  roots  of  unity 116 

109.  Fifth  roots  of  unity        .        .        .        ...        .        .117 

110.  Square  root  of  a  complex  number 118 

111.  Any  root  of  a  complex  number     ......  119 

112.  Sin  n  a  and  cos  n  a.  expressed  in  terms  of  sin  a  and  cos  a    .  120 

113.  Comparison  of  the  values  of  sin  a,  a,  and  tan  a,  a  being 

any  acute  angle 121 

114.  Value  of  sm  a  for  small  values  of  a 121 

a 

115.  Sine  and  cosine  series 122 

116.  Examples 124 


SPHERICAL   TRIGONOMETRY 
CHAPTER  X 

FUNDAMENTAL  FORMULAS 

117.  The  spherical  triangle 127 

118.  Law  of  sines 128 

119.  Law  of  cosines 129 

120.  Law  of  cosines  extended 130 

121.  Relation  between  one  side  and  three  angles         .        .        .  131 

122.  The  sine-cosine  law  131 


CONTENTS 


123.  Relation  between  two  sides  and  three  angles       .        .        .     132 

124.  Relation  between  two  sides  and  two  angles  .        .        .        .132 
126.   Formulas  independent  of  the  radius  of  the  sphere       .        .     133 


CHAPTER  XI 

SPHERICAL  RIGHT  TRIANGLE 

126.  Definition  of  spherical  right  triangle 134 

127.  Formulas  for  the  solution  of  right  triangles  ....  134 

128.  Direct  geometric  derivation  of  formulas       ....  134 

129.  Sufficiency  of  formulas 137 

130.  Comparison   of  formulas    of    plane    and    spherical    right 

triangles 137 

131.  Napier's  rules 138 

132.  Side  and  angle  opposite  terminate  in  same  quadrant  .        .  139 

133.  Two  sides  determine  quadrant  of  a  third      ....  139 

134.  Two  parts  determine  a  triangle 139 

135.  The  quadrant  of  any  required  part 140 

136.  Check  formula 140 

137.  Solution  of  a  right  triangle 141 

138.  Two  solutions 141 

139.  Examples 143 

140.  Quadrantal  triangles 143 

141.  Isosceles  triangles 143 

142.  Examples 144 

CHAPTER  XII 

OBLIQUE   SPHERICAL   TRIANGLE 

143.  Statement  of  aim  of  chapter 145 

GENERAL  SOLUTION 

144.  Angles  determined  from  the  three  sides       ....  145 

145.  Sides  found  from  the  three  angles 146 

146.  Delambre's  or  Gauss's  formulas    .        .        .        .        .        .147 

147.  Napier's  analogies 147 

148.  Formulas  collected        .        . 148 

149.  All  formulas  excepting  law  of  sines  determine  quadrant      .  148 


CONTENTS  xi 


ART.  *AGE 

150.  Theorem  to  determine  quadrant  ......  149 

151.  Second  theorem  to  determine  quadrant        ....  149 

152.  Third  theorem  to  determine  quadrant  .....  150 

153.  Illustrative  examples     ........  150 

154.  Two  solutions        .........  153 

155.  Area  of  spherical  triangle      .......  154 

156.  Examples       ..........  154 


SOLUTION  WHEN  ONLY  ONE  PART  is  REQUIRED 

157.  Statement  of  problem    .......        .156 

158.  From  two  sides  and  the  included  angle,  to  find  any  one  of 

the  remaining  parts        .        .         .         .        .     '   .        .  156 

169.   Two  parts  required        ........  158 

160.  Problems        ..........  159 

161.  Each  unknown  part  found  from  two  angles  and  the  included 

side         ..........     160 

162.  Each  unknown  part  found  from  two  sides  and  an  angle 

opposite  one  of  them      .......     161 

163.  Each  unknown  part  found  from  two  angles  and  a  side 

opposite  one  of  them      .......     163 

164.  The  general  triangle       ........     164 

ANSWERS         ...........    165 


PLANE    TRIGONOMETRY 

CHAPTER   I 
RECTANGULAR   COORDINATES   AND   ANGLES 

1.  Introduction.     The  word  trigonometry  is  derived  from 
the  two  Greek  words  for  triangle  (r/otytuvov)  and  measure- 
ment   (/Aerpta).      Originally    trigonometry    was    concerned 
chiefly  with  the  solution  of  triangles.     At  present  this  is 
but  one  part  of  the  subject. 

Certain  preliminary  considerations,  concerning  directed 
lines  and  angles,  are  neqessary  before  introducing  the  fun- 
damental definitions  of  trigonometry. 

RECTANGULAR  COORDINATES 

2.  Directed  lines.     A  positive  and  a  negative   direction 
may  be  assigned  arbitrarily  to  every  line. 

If  the  direction  from  A  to  C  is  positive,  the  opposite 
direction  from  C  to  A  is  negative. 

If  we  let  the  order  of  the  let-  ^ 

ters   indicate   the   direction  in          A  c 

which  a  segment  is  measured,  it  is  evident  that  AC  and  CA 
represent  the  same  segment  measured  in  opposite  direc- 
tions; hence 

AC  =  -  CA  or  -AC  =  CA. 

Also,  if  B  be  a  third  point  on  the  line,  the  segments  AB 
and  BA  are  opposite  in  sign;  likewise  the  segments  BO 
and  CB.  Hence 

AB=-BA  or  -AB  =  BA, 
and  BC=  -  CB  or  -  BC=  CB. 

1 


2  TRIGONOMETRY 

Then  for  all  positions  of  A,  B,  and  C  on  a  line  it  follows 
that 


Thus  for  the  following  figures  : 
%         K_i_f       Ti     P»                        Q 

5 

8 

12        8  +  f       4^         12 

-12 

A     -8 

C 

B 

0 

4-_6  +  (4-l(r)-4 

C-€ 

B 

0 

4 

\  A 

X' 


3.  Lines  of  reference.     Two  directed  lines,  perpendicular 
to  each  other,  may  be  taken  as  lines  of  reference  or  axes. 
They  are   usually  designated   by    X'X  and   F'F  and  are 
called  the  X-axis  and  the  F-axis  respectively. 

The  X-axis  is  positive  from  left 
to  right  and  negative  from  right 
to  left. 

The  F-axis  is  positive  upward 
_j_       and  negative  downward. 
X  The  point  of  intersection  of  the 

axes  is  the  origin.  The  origin 
serves  as  a  convenient  starting 
point  from  which  to  measure  dis- 
tances. 

4.  Quadrants.     The  axes  produced  divide  the  plane  into 
four  parts  called  quadrants.     The  quadrants  are  designated 
by  number.     The  first  quadrant  is  indicated  by  XOF,  the 
second  by  FOX',  the  third  by  X'OF',  and  the  fourth  by 
F'OX. 

5.  Coordinates  of  a  point.      From  any  given  point  Px  in 
the  plane,  draw  a  line  parallel  to  the  F-axis  intersecting 
the  X-axis  in  some  point  A. 

Then  the  distance  from  the  origin  to  the  point  of  inter- 
section, or  OA,  is  the  abscissa  of  the  given  point  P±. 


RECTANGULAR  COORDINATES  AND  ANGLES       3 


X'C 


The  distance  from  the 
point  of  intersection  to 
the  given  point,  or  APlf 
is  the  ordinate  of  the 
given  point  Px. 

The  abscissa  and  the 
ordinate  of  the  point  P2 
are  OC  and  CP2  respec- 
tively. 

The  abscissa  and  the 
ordinate  of  the  point  P3 
are  OC  and  CP3  respectively. 

The  abscissa  and  the  ordinate  of  the  point  P4  are  OA  and 
AP4  respectively. 

6.  Signs   of   coordinates.      Abscissas   are    positive   when 
measured  from  the  origin  to  the  right,  and  negative  when 
measured  from  the  origin  to  the  left. 

Ordinates  are  positive  when  measured  upward  from  the 
.X-axis  and  negative  when  measured  downward. 

Thus  OA,  APv  and  CP2  are  positive;  0(7,  CP3,  and  AP4 
are  negative. 

7.  EXERCISES 

1.  Locate  the  point  whose  abscissa  is  4  and  whose  ordi- 
nate is  7.     This  point  is  designated  (4,  7). 

2.  Locate  the  point  whose  abscissa  is  2  and  whose  ordi- 
nate is  5,  i.e.  the  point  (2,  5). 

3.  Locate  the  points  (-  3,  4),  (-  6,  -  3),  (5,  -  1),  (0,  4), 
(-  7,  0),  and  (-  8,  10). 

4.  Locate  the  points  (1J,  3),  (-  |,  0),  (m,  n),  (a?,  y),  (a,  0). 

5.  What  is  the  locus  of  the  points  whose  abscissas  are  6  ? 

6.  What  is  the  locus  of  the  points  whose  ordinates  are  —  3  ? 

7.  What  is  the  locus  of  the  points  whose  abscissas  are 
twice  their  ordinates  ? 


TRIGONOMETRY 


ANGLES 

8.  Magnitude   of  angles.      In   elementary   geometry  the 
angles  considered  are  usually  less  than  two  right  angles ; 

but  in  trigonometry  it  is  necessary 
to  introduce  angles  of  any  magni- 
tude, positive  or  negative. 

To  extend   the   conception  of  an 
angle,  suppose  a  line  to  revolve  in 
a  fixed  plane  about  a  fixed  point  0, 
from  the  initial  position  OX  to  the 
successive  positions   OY}  OX',  and 
OP,  generating  a*,  an  angle  greater 
than  two  right  angles. 
•  If  the   line  continues  to   revolve, 
making  more  than  one  complete  rev- 
olution, it  generates  an  angle  a  which 
is    greater   than   four   right   angles. 
Evidently  by  continuing  the  rotation 
an  angle  of  any  magnitude  may  be 
generated.     Thus  the  size  of  the  angle  a  depends  upon  the 
amount  of  rotation  of  OP  and  is  designated  by  an  arc. 

9.  Direction  of  rotation.     Positive  and  negative  angles.     As 
a  positive  and  negative  direction  may  be  assigned  arbitra- 
rily to  a  line,  so  a   positive  and  nega- 
tive sense  of  generation  may  be  assigned 
arbitrarily  to  an  angle. 

The  direction  of  rotation  indicated 
—  by  the  arrows  in  the  figures  of  Art.  8 
is  the  positive  direction ;  the  opposite 
direction,  indicated  by  the  arrow  in  the  adjoining  figure,  is 
the  negative  direction. 

*  Angles  will  usually  be  designated  by  Greek  letters : 

a  ...  Alpha  5  ...  Delta 

/3  .     .     .  Beta  6  .     .     .  Theta 

7  ...  Gamma  0  ...  Phi 


RECTANGULAR  COORDINATES  AND  ANGLES       5 

A  positive  angle  is  an  angle  generated  by  a  line  rotating 
in  the  positive  direction. 

A  negative  angle  is  an  angle  generated  by  a  line  rotating 
in  the  negative  direction. 

Negative  angles,  like  positive  angles,  may  have  unlimited 
magnitude. 

10.  Initial  and  terminal  lines.     The  fixed  line  from  which 
both  positive  and  negative  angles  are  measured  is  the  initial 
line.      It  usually  coincides  with  that   part  of   the   X-axis 
lying  to  the  right  of  the  origin.     Thus,  in  the  last  three 
figures,  OX  is  the  initial  line. 

The  final  position  of  the  revolving  line,  marking  the  ter- 
mination of  the  angle,  is  the  terminal  line.  Thus,  in  the 
last  three  figures,  OP  is  the  terminal  line. 

Two  or  more  unequal  angles  may  PN 
have  the  same  terminal  line.     Thus 
the  positive   angle  a  and  the  nega- 
tive angles  ft  and  y  have  the  same 
terminal  line. 

Angles  having  the  same  terminal 
line,  and  the  same  initial  line,  are  called  coterminal  angles. 

11.  Sign  of  terminal  line.     The  terminal  line,  drawn  from 

Q  f  the  origin  in  the  direction  of  the  ex- 

/  tremity  of  the  measuring  arc,  is  posi- 

/  tive ;     the    terminal     line     produced, 

drawn   from   the   origin  in  the  oppo- 
X    site  direction,  is  negative. 

Thus  the  terminal  line  OP,  of   the 
,  angle  a,  is  positive,  and  the  terminal 

line   produced,   OQ,  is  negative.      As 
the  terminal  line  OP  revolves  it  retains  its  positive  sign. 

12.  The  algebraic  sum  of  two  angles.      To  construct  the 
algebraic  sum  of  two  angles,  conceive  OP  to  rotate  from  the 


6 


TRIGONOMETRY 


position  OX}  through  the  angle  a,  to  OP';   then  from  this 


Pf 


0  X 

a  positive 
R  positive 


O  X 

a  positive 
R  negative. 


a  negative 
positive 

position  let  it  rotate  through  the  angle  fi,  whether  positive 
or  negative,  to  OP.     Then  XOP  is  the  desired  angle  a  +  ft. 

13.  Measurement  of  angles.  Various  units  may  be  em- 
ployed in  the  measurement  of  angles.  In  elementary  geom- 
etry the  right  angle  is  frequently  used.  Two  other  units  in 
common  use  are  the  degree  and  the  radian.  The  degree 
is  generally  used  in  practical  problems  involving  numerical 
computations,  while  the  radian  is  essential  in  many  theoret- 
ical considerations. 

The  degree  is  defined  as  one  ninetieth  of  a  right  angle. 
The  degree  is  divided  into  sixty  equal  parts  called  min- 
utes. The  minute  is  divided  into  sixty  equal  parts  called 
seconds. 

Then  60  seconds  (60")  =  1  minute. 

60  minutes  (60')  =  1  degree. 
90  degrees  (90°)  =  1  right  angle. 

The  angle  26  degrees,  39  minutes,  and  57  seconds  is 
written  26°  39'  57". 


RECTANGULAR  COORDINATES   AND   ANGLES       7 

14.  Circular  or  radian  measure.     A  radian  is  an  angle  of 
such  magnitude  that,  if  placed  with  its  vertex  at  the  center 
of  any  circle,  it  will  intercept  an  arc  equal 

in  length  to  the  radius  of  the  circle. 

Thus,  if  the  arc  XP  is  equal  to  the 
radius  OX,  the  angle  XOP  is  a  radian, 
or  Z.  XOP  =  lr,  r  being  used  to  designate 
radian. 

15.  Value  of  the  radian.     Since,  in  the  same  circle  or  in 
equal  circles,  angles  at  the  center  are  proportional  to  their 
intercepted  arcs,  it  follows  that 

/.XOP 
ZXOY 

one  radian 


or 


one  right  angle      i  (2  irr) 


2 

Therefore,  one  radian  =  —  •  one  right  angle. 

7T 

or  TT  radians  =  180°. 

It  is  clear  that  the  value  of  the  radian  is  independent  of 
the  radius,  depending  only  upon  the  constant  TT  and  the 
right  angle,  and  hence  is  an  invariable  unit. 

16.  Relation  between  degree  and  radian.  From  the  pre- 
ceding article  it  follows  that 

i-=^,  a) 


' 

Also,  from  equation  (1), 

r±jf_  (3) 

180' 
or  1°  =  0.01745''.  (4) 

Equations  (1)  and  (2)  are  used  to  convert  radians  into 


8  TRIGONOMETRY 

degrees,  and   equations   (3)  and  (4)  are   used  to   convert 
degrees  into  radians.     Thus, 


fromd), 

o  Y   TT 

from  (2),  4r  =  4  (57°  17'  44")  =  229°  10'  56", 

from(3),  20°= 

from  (4),  3°  =  3  (0.01745')  =  0.05235r. 

Using  equation  (3)  to  convert  degrees  into  radians  intro- 
duces TT  into  the  numerical  value  of  the  angle.  Thus  TT 
becomes  associated  with  radian  measure.  Since  the  radian 
is  the  angular  unit  with  which  TT  is  commonly  associated, 
no  ambiguity  arises  by  omitting  to  state  with  each  angle, 
expressed  in  terms  of  TT,  that  the  radian  is  the  unit. 

Thus,  90°  =  -  radians,  180°  =  TT  radians,  29°  =  radians 
are  usually  written 

90°  =  -,  180°  =  TT,  29°  = 

2  loO 

It  must  be  especially  noted  that  when  no  unit  is  specified 
the  radian  is  always  understood.  The  constant  TT  is  always 
equal  to  3.1416,  and  can  never  equal  180°,  but  TT  radians 
are  equal  to  180°. 

17.  Relation  between  angle,  radius,  and  arc.  It  is  evident 
from  the  definition  of  a  radian  that  if 
any  arc  of  a  circle  AB  be  divided  by 

^^  the  radius,  the  quotient  indicates  the 

number  of  radians  contained  in  the 
central  angle  subtended  by  the  given 
arc,  hence 


radius 
where  angle  AOB  is  expressed  in  radians.     Representing 


RECTANGULAR  COORDINATES  AND  ANGLES       9 

the  length  of  the  arc  AB  by  a,  the  radius  by  r,  and  the 
angle  AOB  by  0,  the  relation  above  may  be  written 


or 

a=r9  (1) 

It  should  be  especially  noted  that  the  angle  0  is  expressed 
in  terms  of  radians. 

In  the  above  figure  9  is  approximately  equal  to  21  radians. 

Equation  (1)  expresses  the  relation  between  a,  r,  and  0, 
and  determines  any  one  of  these  elements  when  the  other 
two  are  known. 

PROBLEM.  What  is  the  length  of  the  arc  subtended  by  a 
central  angle  of  112°  in  a  circle  whose  radius  is  15  feet  ? 

Solution.     Reducing  the  angle  to  radians  it  is  seen  that 

112°  =?^-r  =  1.95'". 
45 

Hence  0  =  1.95. 

Substituting  in  Equation  (1)  we  have 

a  =  15  x  1.95  =  29.3 
Hence  the  length  of  the  arc  is  29.3  ft. 

18.  Linear  and  angular  velocity.  Equation  (1)  of  Art.  17 
leads  directly  to  the  relation  between  linear  velocity  and 
angular  velocity  in  uniform  motion  in  a  circle.  Suppose  a 
point  P  moves  along  the  circumference  of  a  circle  at  a 
constant  velocity  v,  describing  an  arc  a  in  time  t  ;  then  a/t 
is  called  the  linear  velocity  and  is  represented  by  v.  During 
the  same  time  t,  the  angle  0  is  generated  ;  and  6/t  is  called 
the  angular  velocity,  and  is  represented  by  the  Greek  letter 
o>  (omega). 


10  TRIGONOMETRY 

Dividing  Equation  (1),  Art.  17,  by  t  gives 
«  =  ^or^  =  r«> 

t  t 

i.e.  the  linear  velocity  is  equal  to  r  times  the  angular  velocity. 

While  in  general  angular  velocity  may  be  expressed  in 
any  units  whatsoever,  in  the  equation  v  =  rco  the  angular 
velocity  must  be  expressed  in  radians  per  unit  of  time. 

PROBLEM  1.  If  a  point  moves  26  feet  in  the  arc  of  a 
circle  of  radius  7  feet  in  3  seconds,  what  is  its  angular 
velocity  ? 

Solution.  The  linear  velocity  of  the  point  is  %£-  feet  per 
second. 

Substituting  in  equation  (1)  we  have 

¥  =  7  a, 
.-.  w  =  |f  =  1.238  radians  per  second. 

PROBLEM  2.  A  flywheel  makes  200  revolutions  per 
minute.  Show  that  its  angular  velocity  is  72,000°  per 

minute  or  —  —  radians  per  second. 

If  the  radius  of  the  flywheel  is  3  feet,  show  that  the 
velocity  of  a  point  on  the  rim  is  42.84  miles  per  hour. 

19.  EXAMPLES 

1.  Construct   the   following   angles :  30°,  45°,  60°,  135°, 
300°,  -  60°,  -  90°,  -  390°,  -  420°. 

2.  Construct    approximately    the    following   angles :    2 
radians,  3|-  radians,  —  ^  radian,  —  4  radians,  9  radians. 

3.  Construct  the  following  angles  : 

7T  7T       7T  5  7T       5  TT 

2'   ~3'  4'1       ~T~'  T' 

4.  Eeduce   the   following  angles  to  radian  or   circular 
measure :    10°,    30°,   45°,   60°,   135°,  225°,    -  270°,    - 12°, 
-  18°,  24°  15',  -  612°  19'  25". 


RECTANGULAR  COORDINATES  AND  ANGLES     11 

5.  Reduce  to  degree   measure :    2  radians,   5   radians, 
—  3  radians,  ^  radian,  —  ^  radian,  b  radians. 

6.  Reduce  to  degrees : 

TT    3_7r    _5jr    3.1416   2    TT  + 1        2 

3'        4    '        "      3    '  7T       "'    TT'  2       '    7T  +  3* 

7.  If  an  arc  of  30  ft.  subtends  an  angle  of  4  radians, 
find  the  radius  of  the  circle. 

8.  In  a  circle  whose  radius  is  5,  the  length  of  an  intercepted 
arc  is  12.     Find  the  angle  (a)  in  radians,  (6)  in  degrees. 

9.  If  two  angles  of  a  plane  triangle  are  respectively  equal 
to  1  radian  and  ^  radian,  express  the  third  angle  in  degrees. 

10.  In  a  circle  whose  radius  is  12  ft.,  find  the  length  of 
the  arc  intercepted  by  a  central  angle  of  16°. 

11.  Find  the  angle  between  the  tangents  to  a  circle  at 
two  points  whose  distance  apart  measured  on  the  arc  of  the 
circle  is  378  ft.,  the  radius  of  the  circle  being  900  ft. 

12.  An    automobile    whose    wheels    are    34    inches    in 
diameter  travels  at  the  rate  of  25  miles  per  hour.     How 
many  revolutions  per  minute  does  a  wheel  make  ?     What  is 
its  angular  velocity  in  radians  per  second  ? 

13.  Assuming  the  earth's  orbit  to  be  a  circle  of  radius 
92,000,000  miles,  what  is  the  velocity  of  the  earth  in  its 
path  in  miles  per  second  ? 

14.  The  rotor  of  a  steam  turbine  is  two  feet  in  diameter 
and  makes  25,000  revolutions  per  minute.     The  blades  of 
the   turbine,  situated   on  the  circumference  of  the   rotor, 
have  one-half  the  velocity  of  the  steam  which  drives  them. 
What  is  the  velocity  of  the  steam  in  feet  per  second  ? 

15.  A  belt  travels  around  two  pulleys  whose  diameters 
are  3  feet  and  10  inches  respectively.     The  larger  pulley 
makes    80    revolutions    per    minute.       Find    the    angular 
velocity  of  the  smaller  pulley  in  radians  per  second,  also 
the  speed  of  the  belt  in  feet  per  minute. 


CHAPTER  II 


TRIGONOMETRIC  FUNCTIONS 

20.  The  trigonometric  functions,  upon  which  trigonom- 
etry is  based,  are  functions  of  an  angle. 

These  functions  are  the  sine,  cosine,  tangent,  cotangent, 
secant,  cosecant,  versed  sine,  and  coversed  sine  of  an  angle. 

For  any  angle  a  they  are  written  sin  a,  cos  a,  tan  a,  cot  a, 
sec  a,  esc  a,  vers  a,  and  covers  a. 

21.  Definitions   of   the  trigonometric  functions.      Let  OX 
and  OP  be  the  initial  and  terminal  lines  respectively  of  any 
angle  a. 

Y 


0 


AX 


X'A 


X    A 


X' 


Let  P  be  any  point  on  the  terminal  line, 

OP  or  r  the  distance  from  the  origin  to  the  point  P, 
OA  or  x  the  abscissa  of  the  point  P,  and 
AP  or  y  the  ordinate  of  the  point  P. 
12 


TRIGONOMETRIC   ^UNCTIONS  13 

It  should  be  noted  that  OP,  OA,  and  AP  are  directed 

lines  and  hence 

r  =  OP  and  not  PO 

x  =  OA  and  not  AO 
y  =  AP  and  not  PA. 

Then  the  trigonometric  functions  are  defined  as  follows : 


sin  a  =  -  = 


ordinate 


r      distance 


r     distance 

Y     ordinate 
tan  a  =  ^  =  —  —  :  — 
x      abscissa 

=  *  =  abscissa 

y      ordinate 
=  /:  =  distance 

x      abscissa 

A*     distance 
esc  a  =  -  =  — 

/     ordinate 

vers  a  =  1  —  cos  a 
covers  a  =  1  —  sin  a 

It  will  be  observed  that  each  of  the  first  six  functions  is 
defined  as  a  ratio  between  two  line  segments.  These  are 
the  fundamental  trigonometric  functions,  and  the  ratios  de- 
fining them  are  called  the  trigonometric  ratios. 

The  first  six  functions  are  called  trigonometric  ratios. 
The  expression  trigonometric  functions  is  more  general  and 
embraces  the  versed  sine,  coversed  sine,  and  the  trigono- 
metric ratios.  It  is  evident,  from  the  definitions,  that  the 
values  of  the  trigonometric  functions  are  abstract  numbers. 

22.   Values  of  the  trigonometric  functions  of  30°,  45°,  and 

606.  A  few  concrete  illustrations  serve  to  show  the  nature 
of  the  trigonometric  functions,  to  fix  ideas,  and  to  prepare 
the  way  for  more  general  considerations. 


14 


TRIGONOMETRY 


Functions  of  30°.  Let  OPF  be  an  equilateral  triangle 
having  its  sides  equal  to  2  units.  Place  the  triangle  with 
a  vertex  at  the  origin  so  that  OX  bisects  the  angle  POP. 
Then,  by  geometry,  the  angle  A  OP  =30°,  the  ordinate 
AP=  1,  and  the  abscissa  OA  =  -\/3. 

Hence,  applying  definitions, 

sin30°  =  i=.500 
P 


cos  30°  =        =.866 


X' 


o 


V3     jA    X 

\j 
F 


V3     3 
cot  30°  =  —  =  V 


1.732 


vers30°=l-iV3=.134     sec  30°  =-^z  =  |V3  =  1.155  • 

covers  30°=  1  -  -  =  i  =  .500    esc  30°  =  |  =  2.000 

2i      2i  J- 

PROBLEM.     Find  the  values  of  the  trigonometric  functions 
of  30°,  as  above,  taking  OP=  1. 

Functions  of  45°.     Let    OAP  be 
a    right-angled     isosceles    triangle 

having  its  sides  OA  and  AP  each    

equal  to  1.    Then  Z.  AOP  =  45°  and     X ' 
the  distance  OP  =  V2. 


Hence,  applying  definitions, 

sin  45°  =  —  =  -  V2 

V2     2 

cos  45°  =  —  =  lV2 

V2     2 

1 


1       A    X 


sec  45°=-^  = 

esc  45°  =  ^—  = 

vers45°  =  l-i- 


cot45°  =    = 


covers  45° 


1-V2 


TRIGONOMETRIC   FUNCTIONS 


15 


PROBLEM.  Find  the  values 
of  the  trigonometric  functions 
of  45°,  taking  OA  =  3. 

Functions  of  60°.     Let  OPFy 

an  equilateral  triangle  having 
each  side  equal  to  2,  be  placed 
as  in  the  figure.  Then  the 
abscissa  and  ordinate  of  P 
are  1  and  V5,  respectively. 
Hence,  applying  definitions, 

sin  60°  = 


1      A          F    X 


cos  60°  =  - 

2 


V3 


2~2 
cot  60°  =  -L  =  |  V3       covers  60°  =  1  -  -  V3 

PROBLEM.  Find  the  values  of  the  functions  of  60°,  as 
above,  taking  OP=  4  a. 

The  values  of  the  sines  and  cosines  of  30°,  45°,  and  60° 
are  used  frequently  and  should  be  memorized.  The  follow- 
ing table  may  be  found  helpful : 

sin  30°  =  i  VI  =  cos  60° 

2 


23.   Values  of  the  trigonometric  functions  of  120°,  135°,  and 

150°.     By  using  the  magnitudes  of  the  figures  of  Art.  22 
and  properly  placing  them  with  respect  to  the  axes,  the 


16 


TRIGONOMETRY 


values  of  the  trigonometric  functions  of  various  angles  may 
be  obtained. 

Functions  of  120°.     From  the  figure  and  definitions  it  is 
evident  that 


V3 


X'      A     -1 


120 


sin  120°  =  -^ 

-: 

2  2 


cos  120°  =  — =  -1 


covers  120°  =  1  — 


V3 


cotl20°  =  -=i  =  --V3 

=  -         sec  120°  =  —  =-2 
2  —  1 

esc  120°=— 7==?  A/3 


Functions  of  135°.     From  the  figure  and  definitions  it  is 
evident  that 


V2     2"                              P 

1            1                                          V2 

:=l  =  _lV2                   i    V 

V2         2 

± 

1                                          X'    A    -1 
_^  ~ 

"l""""1                                           Y 

0               X. 

sec  135°  =        =  -  V2 
esc  135°  =  -^?=  V2 


covers  135°  =  l- 


TRIGONOMETRIC   FUNCTIONS 


IT 


Functions  of  150°.     From  the  figure  and  definitions  it  is 
evident  that 


sin  150°  =  - 


cos  150°  = 
tan  150°  = 

cot  150°  = 
sec  150°  = 


-V3 

-V3: 


-V3 


X      A    -Y3 


3V 
V3 


?V3 
3 


J50" 


csc  150°  =    = 


covers  150°  =  l-    = 

w  w 


PROBLEM.  Find  the  values  of  the  trigonometric  functions 
of  210°,  225°,  240°,  300°,  315°,  and  330°. 

24.  Signs  of  the  trigonometric  ratios.  The  signs  of  the 
trigonometric  ratios  of  any  angle  depend  upon  the  signs 
of  the  ordinate,  the  abscissa,  and  the  distance  of  any  point 
on  its  terminal  line.  As  the  terminal  line  passes  from 
one  quadrant  to  another,  there  is  always  a  change  of 
sign  in  either  the  abscissa  or  the  ordinate  of  any  point 
on  that  line,  but  the  distance  remains  positive.  When  a 
coordinate  changes  its  sign,  every  trigonometric  ratio  de- 
pendent upon  it  must  also  change  its  sign. 

The  following  table  is  constructed  by  taking  account  of 
the  signs  of  the  abscissa  x  and  of  the  ordinate  y,  and  re- 
membering that  the  distance  r  is  always  positive.  It  gives 
the  sign  of  each  trigonometric  ratio  of  an  angle  terminating 
in  any  quadrant. 


18 


TRIGONOMETRY 


1st  Quadrant 

2d  Quadrant 

3d  Quadrant 

4th  Quadrant 

-t- 

+  . 

— 

— 

+  -"*• 

+  ~ 

+ 

+ 

cos  a 

±  =  + 

T= 

—  =  — 

jH 

tan  a 

H 

±_ 

3=4- 

T= 

cot  a 

|^ 

+  ~ 

3=  + 

4-  _ 

sec  a 

i^ 

+  _ 

±_ 

1^ 

pep    /y 

+. 

+  . 

+ 

+ 

+- 

Hh 

— 

— 

A2  A 


25.   Theorem.     For  every  given  angle  there  is  one  and  only 
one  value  of  each  trigonometric  function. 

The  theorem  is  demonstrated 
for  the  sine  of  an  angle.     The 
same  method  is  applicable  to 
each  of  the  remaining  functions. 
-         Let  a  be  any  angle.     Eefer- 
ring  to  Art.  21,  it  is  clear  that 
if  it  be  possible  to  obtain  two 
or  more  values  for  sin  a  they 
must  be  obtained  by  taking  dif- 
ferent points  on  the  terminal  line. 

Let  Pj  and  P2  be  any  two  points  on  the  terminal  line. 
Then  by  definition 


Xf 


0 


But  since  the  right  triangles  OA^  and  OA2P3  are  similar, 


TRIGONOMETRIC   FUNCTIONS 


19 


and  have  their  corresponding  sides  in  the  same  direction,  it 
follows  that 


OP,      OP2 

Hence  the  value  of  sin  a  is  independent  of  the  position  of 
the  point  chosen  on  the  terminal  line,  but  depends  solely 
upon  the  position  of  the  terminal  line,  i.e.  upon  the  angle. 

The  above  theorem  may  be  stated  as  follows  :  The  trigo- 
nometric functions  are  single-valued  functions  of  the  angle. 

26.  Theorem.  Every  given  value  of  a  trigonometric  func- 
tion determines  an  unlimited  or  infinite  number  of  positive, 
and  negative  angles,  among  which  there  are  always  two  positive 
angles  less  than  360°. 

The  theorem  is  demonstrated  for  a  given  tangent.  A 
similar  method  is  applicable  to  the  remaining  functions. 

Let  tan  a  =  —  —  where  m 
n 

and  n  are  positive  numbers. 
Then 

-f-  m      —  m 


tan  a  — 


—  n 


Locate  the  point  Pl5  whose 
abscissa  is  —  n,  and  whose 
ordinate  is  m.  This  point 
and  the  origin  determine  the  terminal  line  of  the  angle  a1} 

whose  tangent  is  —  —  • 
n 

Likewise  the  point  P2  is  located  by  using  —  m  and  n  as 
ordinate  and  abscissa  respectively.  Drawing  the  terminal 
line  OP2,  a  second  angle  a2  is  found,  which  also  has  the 
given  tangent. 

The  angles  a,  and  <*2>  are  evidently  less  than  360°,  and 

have  the  given  tangent  —  —  •  There  is  an  unlimited  num- 
ber of  positive  and  negative  angles  coterminal  with  a,  and 
«2,  all  of  which  have  the  given  tangent.  Hence  the  theorem. 


20 


TRIGONOMETRY 


27.  EXAMPLES 

In  what  quadrants  does  a  terminate  when 

1.  sin  a  =  —  i          3.    tan  a  =  5.  5.    sin  a  =  £. 

L  <j 

2.  cos  a  =  i.  4.   cot  a  =  —  8.       "^  6.   cos  a  =  —  4. 

o  5 

7.  sin  a  is  positive  and  cos  «  is  negative. 

8.  tan  ct  is  positive  and  cos  a  is  negative. 

9.  cosec  a  is  negative  and  cos  a  is  negative. 

10.  tan  a  is  negative  and  sin  a  is  positive. 

11.  cos  a  is  negative  and  sin  «  is  negative. 

Give  the  signs  of  the  trigonometric  functions  of  the  fol- 
lowing angles : 


12.   750°. 

i*=. 


U14.   560°. 

^    15.      5|7T. 


16.  -15°. 

17.  -  — • 


18.   -470°. 
10         77r 

"T 


Find  the  negative  angles,  numerically  less  than  360°,  that 
are  coterminal  with  the  following  angles : 


20.  ±^ 

3 

21.  *£. 


22.    300°. 


23. 


24.      5. 


25.    -495° 


26.  Construct  the  positive  angles,  less  than  360°,  for  which 
the  sine  is  equal  to  J- ,  and  find  the  values  of  the  other  func- 
tions of  both  angles. 

SOLUTION.  Determine  the  points 
whose  ordinates  are  2  and  whose 
distances  are  5,  as  follows : 

With  0  as  center  and  a  radius  5, 
describe  a  circle.  Through  a  point 
on  the  Y"-axis,  2  units  above  the 
origin,  draw  a  line  parallel  to  the 
X-axis,  intersecting  the  circle  in 
the  two  required  points  PI  and  P2. 
Draw  the  terminal  lines  OPi  and 
OP»,  giving  the  angles  «i  and  #2,  f°* 
which  we  have 


TRIGONOMETRIC   FUNCTIONS  21 


2  2 

sin  a,i  =  -  and        sin  «2  =  - 

5  5 


-V2T 

COS  «i  =  -  COS  «2  =  - 


tan  0!  =  -      = 


\/21      21  -V21          21 


5  5 

CSC  «i  =  -  CSC  «2  =  ~ 

27.  Construct  the  positive  angles,  less  than  360°,  for 
which  the  cosine  is  f,  and  find  the  values  of  the  other  func- 
tions of  both  angles. 

Find  the  values  of  the  functions  of  all  angles  less  than 
360°  determined  by 

^  28.  tana  =  -f  ^32.  sin  «  =  -•}. 

29.  cot  a  =  If  33.  sin<*  =  ^. 

30.  esc  a  =  3J*  34.  cos  a  =  A. 
-31.  cota  =  -5. 

35.    Given  tan  a  =  -  4,  find  the  value  of  sm  a  cos  a 


36.    Given  sec  «  =  6;  find  the  value  of 


cot  a 
sin2  a  +  cos2  a 


cos  a 

37.  Given  sin  a  =  .3,  find  the  value  of  tan  a  sec  a  cos  a. 

38.  Given   esc  a  =  8   and   tan  ft  =  3,  find   the   value   of 
sin  a  cos  ft  -f  cos  a  sin  ft. 

39.  Given  tan«  =  i  and  cos/3  =  —  ^-VS,  a  terminating 
in  the  first  quadrant  and  /3  in  the  second,  show  that  the  angle 
between  the  terminal  lines  of  a  and  (3  is  a  right  angle. 


CHAPTER   III 


RIGHT    TRIANGLES 

\ 

28.  In  every  triangle  there  are    six   elements   or   parts. 
These  are  the  three  sides  and  the  three  angles. 

When  three  elements  are  given,  one  of  which  is  a  side, 
the  other  three  elements  can  be  determined. 

The  solution  of  a  triangle  is  the  process  of  determining 
the  unknown  parts  from  the  given  parts. 

In  the  present  chapter  it  will  be  shown  how  the  trigono- 
metric functions  can  be  employed  to  solve  the  right  triangle. 

29.  Applications   of  definitions  of  the  trigonometric  func- 
tions to   the   right  triangle.     Let  ABC  be   any  right   tri- 
angle.    Place  the  triangle  in  the  first  quadrant  with  the 
vertex   of  one  acute  angle  coinciding  with  the  origin,  and 
one  side,  not  the  hypotenuse,  coinciding  with  the  X-axis,  as 
in  the  figure. 

Y 


B 


X' 


Then,  by  definitions, 


sin  a  =  -  = 


side  opposite 
hypotenuse 
side  adjacent 


cos  a  =  -  = 

c       hypotenuse 

22 


RIGHT   TRIANGLES  23 

=  g  =  side  opposite 
b      side  adjacent 

6      side  adjacent 
cot  a  =  -  =  — 

a     side  opposite 

==c=  hypotenuse 
6      side  adjacent 


a     side  opposite 

It  is  seen  that  the  functions  of  any  acute  angle  of  a  right 
triangle  are  expressed  in  terms  of  the  side  adjacent,  the 
side  opposite,  and  the  hypotenuse,  without  reference  to  the 
coordinate  axes. 

It  follows  that 

sin/3  =  -  cot0  =  - 

C  0 

cos/?  =  -  sec/3  =  - 

c  a 


30.  Trigonometric  tables.  In  the  preceding  chapter  the 
trigonometric  functions  of  30°,  45°,  and  60°  were  calculated. 
By  processes  too  complicated  to  introduce  here,  tables  have 
been  computed,  giving  the  values  of  the  trigonometric  func- 
tions of  acute  angles.  These  tables  generally  contain  two 
parts.  In  one  part  the  values  of  the  functions,  called 
natural  functions,  are  given  ;  in  the  other  part  the  logarithms 
of  the  trigonometric  functions,  called  logarithmic  functions, 
are  given. 

When  an  angle  is  given  its  trigonometric  functions  can 
be  taken  from  the  tables,  and  vice  versa.  The  functions  of 
known  angles  thus  become  known  numbers  and  can  be  used 
in  problems  of  computation. 

Approximate  values  of  the  trigonometric  functions  can  be 
obtained  by  graphical  methods. 


24  TRIGONOMETRY 

PROBLEM.  Measure  the  distance,  abscissas  and  ordinates 
of  the  points  a,  6,  c,  •  •  •  ,  j.  From  these  measurements 
compute  to  two  figures  the  sine,  cosine,  and  tangent  of  0°, 
10°,  20°,  •  .  •  ,  90°.  By  arranging  the  results  in  tabular 
form,  a  two  place  table  is  constructed. 


70C 


60< 


,IOe 


This  table  may  be  used  to  solve  Examples  1  to  6,  Art.  36. 
On  account  of  its  extreme  simplicity  the  use  of  this  table 
allows  the  attention  to  be  focused  upon  the  fundamental 
processes  involved  in  the  solution  of  triangles. 


lno 


RIGHT  TRIANGLES  25 

A 
31.   Formulas  used  in  the  solution  of  right  triangles.     A 

right  triangle  can  always  be  solved  when,  in  addition  to 
the  right  angle  y,  two  independent  parts  are  given.  The 
formulas  usually  employed  are  : 


c  a2  +  62  =  c2 

cos  a  =  -  cos  8  =  - 

c  c 

tona=-  tan/?  =  - 

o  a 

When  the  computations  are  made  without  logarithms, 
the  formulas  involving  the  cotangent,  secant,  and  cosecant 
may  sometimes  be  used  advantageously. 

32.  Selection  of  formulas. 

(a)  If  an  angle  and  a  side  are  given,  it  is  always  possible 
to  find  the  unknown  parts  directly  from  the  given  parts 
without  the  use  of  the  formula  a2  -f-  62  =  c2.  To  find  the 
unknown  side,  select  that  formula  which  contains  the  given 
parts  and  the  desired  unknown  side.  The  unknown  angle 
can  always  be  found  from  a.  -f  /?  =  90°. 

(6)  If  two  sides  are  given,  the  third  side  would  naturally 
be  found  by  the  use  of  a2  +  b2  =  c2,  but  in  practice  it  is 
generally  preferable  first  to  compute  an  angle  by  the  use  of 
a  formula  involving  the  given  sides  and  an  angle.  To  find 
the  third  side,  select  a  formula  containing  one  of  the  given 
sides,  the  angle  already  computed,  and  the  required  side. 

33.  Check  formulas.     In  all  computations  it  is  necessary 
constantly  to   guard  against  numerical   errors.     However 
carefully  the  computations  are  made,  errors  may  still  occur, 
and  therefore  computed  parts  should  be  checked  by  means 
of  check  formulas.     Any  formula  which  was  not  used  in 
the  solution  of  the  triangle  may  be  used  for  this  purpose. 

For  the  right  triangle  the  formula 

may  conveniently  be  used,  and  in  general  it  is  sufficient. 


26  TRIGONOMETRY 

34.  Suggestions   on  solving   a  triangle.     Make  a  careful 
free-hand  construction  of  the  required  triangle,  and  write 
down   an   estimate   of   the   values  of   the  unknown  parts. 
Large  errors  will  be  detected  readily,  without  the  use  of 
check   formulas,  when  the  computed  parts   are   compared 
with  the  estimates. 

Before  entering  the  tables,  and  before  making  any  com- 
putations, select  all  the  formulas  to  be  used,  solve  the 
formulas  for  the  required  parts,  and  make  an  outline  in 
which  a  place  is  provided  for  every  number  to  be  used  in 
the  computation.  This  will  often  lessen  the  actual  work, 
for  frequently  several  required  numbers  are  found  on  the 
same  page  of  the  table. 

The  arrangement  of  the  work  is  of  considerable  impor- 
tance in  every  extended  computation. 

35.  Illustrative  examples. 

1.  In  a  right  triangle,  given  b  =  14,  a  =  35°,  to  find  a,  c, 
and  ft.  ^ 

SOLUTION.     Approximate  construction.  S^<\ 

Estimate  a  =  9,  c  =  17.  %,'''  \a 

By  natural  functions 


a--              cosa  =  - 

0                                         C 
—  7)  tan  flf           '    r  —                              ft  —  Q0°        ct. 

6=14 

Check 
c2  =  a2  +  &2 
a2=    96.10 
62  =  196.0 

cos  a 
a  =  14  x.  7002       c  =  14  -*-  .8192    j8  =  56° 
a  =  9.803               c  =  17.09 

By  logarithms 


292.1 
c2  =  (17.09)2  =  292.1 


a  =  6tana  c  =  —  0  =  90°  -  « 


b 
a 

log  6 
log  tan  a 

log  a 
a 


14  log  b 

35°  log  cos  a 

logo 


c 


c 

6 
c-b 

c+  b 


RIGHT   TRIANGLES 

Check 

6) 


log  (c  -  6) 

log  (c  +  6) 

2  log  a 

log  a 


Filling  in  the  above  outline,  the  completed  work  appears  as  follows ; 

b  0  nno 


a  =  b  tan  a 


b 
a 
log  6 
log  tan  a 
log  a 
a 

14 
35° 
1.14613 
9.84523 

-10 

0.99136 
9.8030 

c  = 


cos  a 


log  b 

1.14613 

0=  55° 

log  cos  a 

9.91336  -  10 

logc 

1.23277 

c 

17.091 

Check 

ai  -  C2  _  52  _  (C  _  5)  (C  +  5^ 

c 

b 
c-b 

c-f  & 

17.091 
14. 

log(c-&) 
log(c  +  &) 

0.49010 
1.49263 

3.091 
31.091 

2  log  a 
log  a 

1.98273 
0.99136 

Since  the  check  formula  gives  the  same  value  for  log  a  as  that  found 
in  the  solution,  the  computation  is  in  all  probability  correct. 

2.    In  a  right  triangle,  given  c  =  6.275,  /3  =  18°47f,  to  find 
a,  &,  and  a. 

SOLUTION.     Approximate  construction. 
Estimated  a  =  5,          b  =  2. 

By  natural  functions  Check 

a  =  c  cos  j8  b  =  c  sin  0  c2  =  a2  +  62 

a  =  6.275  x  .9468      b  =  6.275  x  .3220  a2  =  35.30 

a  =  5.941  b  =  2.021  62  =    4.084 

a  =  90°  -  ft  39.384 

a  =  71°  13'          c2  =  (6.275)2  =  39.38 


28 


TRIGONOMETRY 


By  logarithms 
a  =  c  cos  /3 

b  =  csin/3 

a  =  90°  -  /a 

c 
ft 

logc 
log  cos  /3 
log  a 
a 

6.275 
18°  47' 
0.79761 
9.97623  -  10 

logc 
log  sin  /3 
log  6 
6 

Check 
2  =  (c  -  6) 

0.79761 
9.50784 

a  =  71°  13' 
-10 

0.30545 
2.0205 

(c+6) 

0.77384 
5.9407 

.      c     6.275 
6     2.0205 

log  (c  -  6) 
log  (c  +  6) 

0.62885 
0.91884 

c  -  &     4.2545 
c  +  6     8.2955 

2  log  a 
log  a 

1.54769 
0.77384 

3.    Given  a  =  .064873,  b  =  .12574,  to 
find  a,  /?,  and  c. 

SOLUTION.     Approximate  construction. 
Estimate    a  —  30°,  /3  -  60°,  c  =  .15. 


sin  a 


a 

.064873 

6 

.12574 

log  a 
log  6 

8.81206  -  10 
9.09948  -  10 

log  tan  a 
a 

9.71258  -  10 
27°  17'  23" 

/3  62°  42'  37" 

c 

.14149 

b 
c-6 

.12574 

.01575 

c  +  6 

.26723 

log  a 

log  sin- a 

logc 

c 


8.81206  -  10 
9.66133  -  10 


9.15073  -  10 
.14149 


Check 


log  (c  -  6) 

log  (c  +  6) 

2  log  a 

log  a 


8.19728-10 
9.42689  -  10 


17.62417  -  20 
8.81208  -  10 


RIGHT   TRIANGLES 


36.  EXAMPLES 

Solve  the  following  right  triangles,  y  being  the  right 
angle.  It  is  recommended  that  the  first  ten  problems  be 
solved  by  the  use  of  a  three-place  table  of  natural  func- 
tions, or  by  means  of  a  slide  rule. 

1.  a  =  6  +A.   c  =  .0091 
a  =  20°  a  =  .0029 

2.  c  =  2.5  7.   6=371 
a  =  35°  a  =  43° 

3.  6  =  .84  8.   c  =  7.72 
ft  =  75°  6  =  6.87 

4.  a  =  25  9.   a=18° 

6  =  60  c  =  .0938 

5.  c  =  82  10.   /?=49°30' 
a  =  37  c  =  12.47 

The  following  problems  should  be  solved  by  the  use  of 
a  four-place  *  or  a  five-place  table. 

11.   a  =  1870  16.    c  =  12.145 

a  =  19°  55'  a  =  9.321 

<**-12.    c=.3194  17.   6  =  78.545 

a  =  25°  41'  «  =  31°  41'  6" 

13.  6  =  .9292  18.   6=3.4572 

/3  =  32°  43'  ft  =  57°  57'  57" 

14.  a  =  .00006  19.   c  =  20.082 
6  =  .000019  6  =  16.174 

15.  c  =  1200.7  20.   a  =  78°  0'  3" 
a  =  885.6  a  =  271.82 

*  Whenever  sides  are  given  to  four  significant  figures  and  angles  to 
minutes. 


30  TRIGONOMETRY 


a  =  5987.2  j6.   /?  =  83°  15'  6" 

ft=  88°  53'  2"  c  =  7000 

22.  c  =  .  09008  27.    a  =66°  6'  18" 
a  =  .07654  c  =  8070.6 

23.  a  =  46°  39'  50"  28.    a  =  978.45 
a  =  26.434  6  =  1067.2 

24.  a  =  30.008  -^29.    a  =  5280 
b  =  29.924  6  =  5608 

25.  a  =  111.45  /A30-    «  =  17°  26'  34" 
b  =  121.69  c  =  46.474 

31.  Solve  the  isosceles  triangle,  one  of  the  equal  sides 
being  690.13,  and  one  of  the  base  angles  being  15°  20'  25". 

32.  Solve  the  isosceles  triangle  whose  altitude  is  606.6, 
one  of  the  equal  sides  being  955.7. 

33.  Solve  the  isosceles  triangle  whose  base  is  2558,  and 
whose  vertical  angle  is  104°  0'  46". 

34.  Solve  the  isosceles  triangle  whose  base  is  161.4,  and 
whose  altitude  is  204.4. 

35.  Find  the  length  of  a  side  of  a  regular  octagon  in- 
scribed in  a  circle  whose  radius  is  49. 

37.  Oblique  triangles.  When  any  three  independent  parts 
of  an  oblique  triangle  are  given  it  can  be  solved  by  means 
of  right  triangles.  The  oblique  triangle  may  be  divided 
into  two  right  triangles  by  drawing  a  perpendicular  from 
one  vertex  to  the  opposite  side,  or  the  opposite  side  pro- 
duced. 

When  one  of  the  given  parts  is  an  angle,  the  perpendicu- 
lar must  be  selected  so  that  one  of  the  resulting  right 
triangles  will  contain  two  of  the  given  parts. 

In  case  the  three  sides  are  given,  a  second  part  of  one  of 


RIGHT   TRIANGLES  31 

the  right  triangles  can  be  obtained  by  equating  the  expres- 
sions for  the  length  of  the  perpendicular  obtained  from  each 
right  triangle. 

Thus     a2  -  a2  =  62  -  (c  -  a)2, 

from  which       a?  =  — 

2c 

38.  APPLICATIONS 

1.  To  find  the  distance  from  B  to  (7,  two  points  on  op- 
posite sides  of  a  river,  a  line  BA,  200  feet  long,  was  laid  off 
at  right  angles  to  the  line  BC,  and  the  angle  BAC  was 
measured  and  found  to  equal  55°  29'.     What  was  the  re- 
quired distance  ? 

2.  A  railway  is   inclined  4°  23 '20"   to   the  horizontal. 
How  many  feet  does  it  rise  per  mile,  measured  along  the 
horizontal  ? 

3.  From  the  top  of  a  tower  120  feet  high  the  angle  of 
depression  of  an  object  in  the  horizontal  plane  of  the  base 
of  the  tower  is  24°  27'.     How  far  is  the  object  from  the  foot 
of  the  tower  ?     How  far  from  the  observer  ? 

Given  any  point  A  and  a  second  point  B  at  a  greater  elevation 
than  A. 

The  angle  of  elevation  of  B  from  A  is  the 
angle  that  the  line  AB  makes  with  its  orthogonal 
projection  upon  the  horizontal  plane  through  A. 
_   In  the  figure  it  is  the  angle  a. 
A  The  angle  of  depression  of  A  from  B  is  the 

angle  that  the  line  BA  makes  with  its  orthogonal  projection  upon  the 
horizontal  plane  through  B.  In  the  figure  it  is  the  angle  /3.  It  is  clear 
that  a  =  0. 

4.  Find  the  length  of  one  side  of  a  regular  pentagon 
inscribed  in  a  circle  whose  radius  is  18.24  feet. 

5.  Find  the  perimeter  of  a  regular  polygon  of  n  sides 
inscribed  in  a  circle  whose  radius  is  r. 


32  TRIGONOMETRY 

6.  Find  the  perimeter  of  a  regular  pentagon  circum- 
scribed about  a  circle  whose  radius  is  18.24  feet. 

7.  Find  the  perimeter  of  a  regular  polygon  of  n  sides 
circumscribed  about  a  circle  whose  radius  is  r. 

8.  Find  the  length  of  a  chord  subtending  a  central  angle 
of  63°  14'  20"  in  a  circle  whose  radius  is  124.93  feet. 

9.  A  straight  road,  PR,  makes  an  angle  of  19°  27'  30" 
with  another  straight  road,  PS.     Having  given  PR  =  640 
feet,  find  US,  the  perpendicular  distance  from  R  to  PS. 

/"*  I 

ACUAt  a  certain  point  the  angle  of  elevation  of  a  moun- 
tain is  34°  28'.  At  a  second  point  500  feet  farther  away, 
the  angle  of  elevation  is  31°  12'.  Find  the  height  of  the 
mountain  above  the  table-land. 

11.  One  side  of  the  square  base  of  a  right  pyramid  is  15 
inches,  and  the  altitude  is  20  inches.     Find  the  slant  height 
and  the  edge.     Find  the  inclination  of  a  face  to  the  base  of 
the  pyramid. 

12.  How  far  can  you  see  from  an  elevation  of  2000  feet, 
assuming  the  earth  to  be  a  sphere  with  a  radius  of  3960 
miles  ? 

13.  Two  forces  of  95.75  pounds  and  120.25  pounds,  at 
right  angles  to  each  other,  act  at  a  point ;  find  the  magnitude 
of  their  resultant  and  the  angle  it  makes  with  the  greater 
force. 

14.  Find   the   velocity   of    a   point,   whose    latitude    is 
44°  30'  20",  due  to  the  rotation  of  the  earth.     Assume  the 
radius  of  the  earth  to  be  3960  miles. 

15.  Find  the  length  of  a  belt  running  around  two  pulleys 
whose  radii  are  12  inches  and  4  inches,  respectively,  the 
distance  between  the  centers  of  the  pulleys  being  6  feet. 

16.  A  diagonal  of  a  cube  and  a  diagonal  of  a  face  of  a 
cube  intersect  at  a  vertex.     Find  the  angle  between  them. 


RIGHT   TRIANGLES  33 

17.  A  force  of  2000  pounds  applied  at  the  origin  makes 
an  angle  of   33°  25'  with  the  positive  X-axis.     Find  its 
components  along  the  X  and  Y  axes  respectively. 

18.  A  force  of  185  pounds  applied  at  the  origin  makes  an 
angle  of  82°  12'  with  the  positive  X-axis.     Another  force  of 
327  pounds  applied  at  the  origin  makes  an  angle  of  11°  32' 
with  the  positive  X-axis.     Find  the  X  and  Y  components 
of  each  of  these  forces.     Add  the  X-components  and  also 
the  y-components  and  then  find  the  resultant  in  magnitude 
and  direction. 

19.  A  cylindrical  tank,  whose  axis  is  horizontal,  is  8  feet 
in  diameter  and  12  feet  long.     The  tank  is  partly  filled  with 
water,  so  that  the  depth  of  the  water  at  the  deepest  point  is 
3  feet.     How  many  gallons  of  water  are  in  the  tank,  there 
being  7-J-  gallons  in  a  cubic  foot  ? 

20.  A  man  who  can  paddle  his  canoe  5  miles  per  hour  in 
still  water  paddles  at  his  usual  rate  directly  across  a  river 
one-half  mile  wide.     If  the  river 'flows  4  miles  per  hour, 
where  will  the  canoe  land  and  what  is  its  speed  in  the  water  ? 

*      '    •*= 


• 


•-A— &-o 
CHAPTER  IV 


O-^v 


VARIATIONS  OF  THE  TRIGONOMETRIC  FUNCTIONS 
REDUCTION   OF   FUNCTIONS   OF  /i90°±a 

39.  In  the  study  of  the  variations  of  the  trigonometric 
functions,  their  values  at  0°,  90°,  ISO0,  270°,  and  360°  are  of 
special  importance. 

Values  of  functions  of  0°.  The  terminal  line  of  0°  coin- 
cides with  the  positive  X-axis.  For  a  point  on  this  line,  at 
a  distance  r  from  the  origin,  we  have 


Y 

x=r, 

y=( 

) 

Hence 

sin 

0°  = 

0 

=  0 

cotO° 

_r*_ 

00 

P 

r 

0~ 

X'              0 

X 
cos 

0°  = 

r 

=  1 

secO° 

_  r  _ 

1 

r 

r 

tan 

0°  = 

0 

=  0 

cscO° 

_  r  _ 

00 

Y 

r 

0 

Values  of  functions  of  90°.      The  terminal  line  of  90°  coin- 
cides  with  the  positive  F-axis,  therefore 


X- 

=  U,         y  =  r. 

Y 

Hence 

P 

sin  90°  =  -  =  1 

cot90°  =  -  =  0 

r 

r 

cos  90°  =  -  =  0 

r             X' 

sec  90°  =.  -  =  oo 

0 

X 

r 

0 

tan90°  =  -  =  oo 

csc90°  =  -  =  l 

i 

0 

r 

Y 

*  For  the 

interpretation  of  -  see  any  College 

Algebra. 

34 

VARIATIONS  OF   TRIGONOMETRIC   FUNCTIONS      35 


Values  of  functions  of  180°.     The  terminal  line  of  180°  co 
incides  with  the   negative   X-axis, 
therefore 


x=  —  r, 


Hence 


sin  180°=   --    =0 
r 

cos  180°  =  —  =  - 1 


X' 


tan  180°  =  — =  0 
—  r 


cot  180°  =  — -  =  oo 


sec  180°  =  — =  -1 
— r 

esc  180°=        =00 


Values  of  functions  of  270°.      The  terminal  line  of  270° 
coincides  with  the  negative  Y-axis,  therefore 

x  =  0,        y  =  —  r 
Hence 

sin  270°  =nr=  -1      cot  270°  =—=0 
r  —r 


cos  270°=  -  =0 
r 


sec  270°=  £  =00 


tan  270°  =—  =00         esc  270°=—  =-1 
0  —  r 

The  values  of  the  functions  of  360°  are  identical  with  the 
values  of  the  functions  of  0°,  since  these  angles  are 
coterminal. 


40.  Variation  of  the  functions.  It  has  been  shown  that 
the  trigonometric  functions  are  functions  of  an  angle.  As 
the  angle  varies,  the  trigonometric  functions  depending  upon 
the  angle  also  vary. 


36 


TRIGONOMETRY 


Let  the  line  OP,  of  fixed 
length  r,  revolve  about  0 
from  the  initial  position  OX. 
Then  the  angle  XOP  or  a 
increases  from  0°  to  360°. 
The  variations  of  the  trigo- 
nometric functions  can  be 
traced  by  observing  the 
changes  in  the  abscissa  and 
ordinate  of  the  point  P. 


a.  As  the  angle  a  increases  from  0°  to  90°, 

y  increases  from  0  to  r, 
and  x  decreases  from  r  to  0. 
Hence 

sin  a  or  -  is  positive  and  increases  from  0  to  1 

fm 

cos  a  or  -  is  positive  and  decreases  from  1  to  0 
tan  a  or  -  is  positive  and  increases  from  0  to  oo 
cot  a  or  -  is  positive  and  decreases  from  co  to  0 

y 

T 

sec  a  or  -  is  positive  and  increases  from  1  to  oo 
esc  a  or  -  is  positive  and  decreases  from  oo  to  1 

y 

As  the  angle  a  increases  through  90°,  x  passes  through 
zero,  changing  from  a  positive  number  to  a  negative  num- 
ber. Then,  immediately  before  a  becomes  90°,  cos  a  or  - 

r 

is  a  very  small  positive  number ;  while  immediately  after 
a  has  passed  90°,  cos  a  is  a  very  small  negative  number. 
This  may  be  expressed  by  saying  that  cos  a  passes  through 
zero  and  changes  sign  as  a  passes  through  90°. 


VARIATIONS   OF   TRIGONOMETRIC   FUNCTIONS      37 

Likewise,  immediately  before  a  becomes  90°,  tan  a  or  £ 

x 

is  a  very  large  positive  number ;  while  immediately  after 
a  has  passed  90°,  tan  a  is  a  very  large  negative  number. 
It  has  been  seen  that  tan  90°  =  oo .  Hence  tan  a  passes 
through  oo  and  changes  sign  as  a  passes  through  90°. 
Hence  we  may  say  that  tan  90°  =  ±  GO  ,  choosing  the  posi- 
tive sign  when  associating  90°  with  the  first  quadrant  and 
the  negative  sign  when  associating  90°  with  the  second 
quadrant. 

Similarly,  whenever  a  trigonometric  function  passes  throiigh 
oo  it  changes  sign. 

b.  As  the  angle  a  increases  from  90°  to  180°, 

y  decreases  from  r  to  0, 
and  x  decreases  from  0  to  —  r. 
Hence 

sin  a  or  ^  is  positive  and  decreases  from  1  to  0 
r 

cos  a  or  -  is  negative  and  decreases  from  0  to  —  1 

tan  a  or  ^  is  negative  and  increases  from  —  oo  to  0 
x 

cot  a  or  -  is  negative  and  decreases  from  0  to  —GO 

y 

sec  a  or  -  is  negative  and  increases  from  —  oo  to  —  1 
x 

esc  a  or  —  is  positive  and  increases  from  1  to  +  oo 

y 

c.   As  the  angle  a  increases  through  180°,  y  passes  through 
zero,  changing  from  a  positive  number  to  a  negative  number. 
As  the  angle  a  increases  from  180°  to  270°, 

y  decreases  from  0  to  —  r, 
and  a  increases  from  —  r  to  0. 
Hence 

sin  a  or  ^  is  negative  and  decreases  from  0  to  —  1 

T 


38  TRIGONOMETRY 


cos  a  or  -  is  negative  and  increases  from  —  1  to  0 

T 

tan  a  or  -  is  positive  and  increases  from  0  to  -f  oo 


cot  a  or  -  is  positive  and  decreases  from  +  oo  to  0 

sec  a  or  -  is  negative  and  decreases  from  —  1  to  —  oo 
x 

esc  a  or  -  is  negative  and  increases  from  —  oo  to  —  1 

d.  As  the  angle  a  increases  through  270°,  x  passes  through 
zero,  changing  from  a  negative  number  to  a  positive 
number. 

As  the  angle  a  increases  from  270°  to  360°, 

y  increases  from  —  r  to  0, 
and  x  increases  from  0  to  r. 
Hence 

sin  a  or  -  is  negative  and  increases  from  —  1  to  0 


flf* 

cos  a  or  -  is  positive  and  increases  from  0  to  1 
tan  «  or  -  is  negative  and  increases  from  —  oo  to  0 


cot  a  or  -  is  negative  and  decreases  from  0  to  —  oo 
sec  a  or  -  is  positive  and  decreases  from  +  oo  to  1 

7* 

esc  a  or  -  is  negative  and  decreases  from  —  1  to  —  oo 
The  above  results  are  presented  in  tabular  form. 


VARIATIONS  OF  TRIGONOMETRIC   FUNCTIONS       39 


( 

1st  QUADRANT  | 

P                           9C 

2d  QUADRANT 
0                         18 

8d  QUADRANT 

3°                        27 

4th  QUADRANT 
0°                   360° 

sin  «=- 

+  0  inc.  -f  1 

+  1  dec.  +  0 

-  0  dec.  -  1 

-  1  inc.  -  0 

X 

cos  a=- 
r 

+  1  dec.  +  0 

-Odec.  -  1 

-  1  inc.  -  0 

+  0  inc.  +  1 

V 

tan  a=- 

X 

-f  0  inc.  +00 

—oo  inc.  —  0 

+  0  inc.  +00 

—ao  inc.  —  0 

cot«=* 

+  00  dec.  +  0 

—  0  dec.  —oo 

+00  dec.  +  0 

-  0  dec.  -oo 

sec  «=- 

X 

+  1  inc.  +00 

—oo  inc.  —  1 

—  1  dec.  —oo 

+  00  dec.  +  1 

esc  a—  - 

y 

+  oo  dec.  +  1 

+  1  inc.  +00 

—  oo  inc.  —  1 

—  1  dec.  —oo 

It  is  thus  seen  that  the  sine  and  cosine  can  never  be 
greater  than  -+- 1  nor  less  than  —  1,  while  the  secant  and 
cosecant  have  no  values  between  -f  1  and  —1,  but  have 
values  ranging  from  +- 1  to  +00  and  from  —  1  to  —  oo  . 

The  tangent  and  cotangent  may 
have  any  value  from  +-  oo  to  —  oo  . 

41,  Graphical  representation.  A 
graphical  representation  of  the  trigo- 
nometric functions  is  effected  by  first 
locating  points  using  the  different 
values  of  the  angle  as  abscissas  and 
the  corresponding  function-values  as 
ordinates,  and  then  drawing  a  smooth 
curve  through  these  points  taken  in 
the  order  of  increasing  angles. 

The  values  of  the  functions  of 
the  angles  previously  calculated  are 
sufficient  to  determine  an  approxi- 
mate graph.  For  greater  accuracy 
the  values  of  the  functions  may  be 
taken  from  the  table  of  natural 
functions. 


a 

since 

a 

tan  « 

0 

0 

0 

0 

7T 

1 

7T 

1 

6~ 

2 

6 

gV3 

it 

-A/~ 

IT 

1 

4 

2 

4 

7T 
3" 

|V3 

7T 
3 

V3' 

7T 

1 

IT 

00 

2 

2 

27T 

1         _ 

2* 

V3 

3 

2 

3 

37T 

1     - 

Sir 

-1 

4 

2 

4 

67T 

1 

STT 

V3 

6 

2 

6 

3 

7T 

0 

IT 

0 

?7T 

1 

77T 

1 

6 

2 

"6~ 

gV3 

etc. 

etc. 

40 


TRIGONOMETRY 

Sine  Curve  y  =  sin  x 


Tangent  Curve  y—tanx 


The  sine  and  tangent  curves  illustrate  the  truth  of  the 
theorems  of  Arts.  25  and  26. 

o 

Thus  for  any  given  value  of  the  angle,  as  —  ^,  there  is  but 

4 

one  value  of  each  function,  the  curves  showing  the  sine  and 
tangent  to  be  1  V2  and  —  1  respectively. 

But  for  any  given  value  of  a  trigonometric  function,  as 
sin  a,  =  ^,  there  are  an  unlimited  number  of  angles,  as 


,     Z,   2J»,   2f»,  etc.  . 

Also  for  tan  a  =  V^we  see  that  a  may  have  the  values 
I    HTT,   21  TT,   31  TT,  etc. 


42.   Periodicity  of  the  trigonometric  functions.     From  the 
sine  curve  it  is  readily  seen  that  the  sine  is  0  when  the 


VARIATIONS  OF   TRIGONOMETRIC   FUNCTIONS      41 

angle  is  0,  and  that  it  increases  to  1  as  the  angle  increases  to 

?-.     At  this  point  the  sine  begins  to  decrease  and  has  the 

2 

value  0  when  the  angle  is  TT,  and  finally  reaches  the  value 

o 

—  1  when  the  angle  is  — - ;  then  the  sine  again  begins  to  in- 

2i 

crease  and  has  the  value  0  when  the  angle  has  the  value  27r. 

When  the  angle  increases  from  2  TT  to  4  TT,  the  sine  repeats 
its  values  of  the  interval  from  0  to  2?r.  If  the  angle  were 
increased  indefinitely,  the  sine  would  repeat  its  values  for 
each  interval  of  2  ?r.  For  this  reason  the  sine  is  called  a 
periodic  function,  and  2  TT  is  its  period. 

A  study  of  the  tangent  curve  shows  that  the  tangent  has 
the  same  values  between  0  and  TT  that  it  has  between  ?r  and 
27T  or  between  2?r  and  3?r.  Hence  the  tangent  is  also  a 
periodic  function  having  the  period- TT. 

43.  EXAMPLES 

1.  Plot  y  =  cos  x  and  give  the  period  of  the  cosine. 

2.  Plot  y  =  cot  x  and  give  the  period  of  the  cotangent. 

3.  Plot  y  =  sec  x  and  give  the  period  of  the  secant. 

4.  Plot  y  =  esc  x  and  give  the  period  of  the  cosecant. 

5.  Plot  y  =  sin  x  +  cos  x. 

6.  Plot  y  =  x  +  sin  x. 


TRIGONOMETRY 


REDUCTION  OF  FUNCTIONS  OF  n  90°  ±  a 

44.  The  formulas  to  be  developed  in  this  section  enable 
us  to  express  any  function  of  any  angle  in  terms  of  a  func- 
tion of  an  angle  differing  from   the   given  angle   by  any 
multiple  of  90°.     They  may  be  used  to  express  any  func- 
tion of  any  angle  in  terms  of  a  function  of  an  angle  less 
than  90°  or  less  than  45°. 

45.  Functions  of  —  a  in  terms  of  functions  of  a(n  =  0). 
Let  XOP  be  any  positive  angle  and  XOP'  a  numerically 
equal  negative  angle. 


Then  taking  OP'  =  OP  we  have,  for  each  figure, 
xf  =  x,  y'  =  -  y,  r'  =  r, 

where  #,  #,  r  and  #',  y',  r'  are  associated  with  P  and  Pf 
respectively. 

Then  in  each  quadrant 


sn   —  a  =  ~f  =     _   —  _  sn  a 

xf     x 
cos  (—«)=  —  =  -  =  cos  a 


VARIATIONS  OF   TRIGONOMETRIC   FUNCTIONS     43 


tan  (—  a)  =  ^-  =  — ^  =  —  tan  a 
x        x 

Xf  X 

cot  (—«)  =  —  = =  —  cot  a 

y'    -y 

r'     r 

sec  (—  a)  —  —  —  -  —  sec  a 
x'     x 

esc  (—  a)  =  —  =  -^-  =  —  esc  a 
2/'      -2/ 

46.  Functions  of  90°  +  a  in  terms  of  functions  of  a(n  =  I), 
Let  XOP  be  any  positive  angle  a  and  XOP'  the  angle 
90°  +  a. 

Then  taking  OP'  =  OP  we  have,  for  each  figure, 


where  a?,  ?/,  r  and  #',  #',  r'  are  associated  with  P  and  Pr 
respectively. 


X' 


x 


Similar  figures  can  be  constructed  for  the  other  quadrants. 
Then  in  each  quadrant 

sin  (90°  +  «)  =  £'  =  -  =  cos  a 


- 
r' 


=  -  snce 


tan  (90°  +  a)  =     =  —  =  -  cot  « 
v     -y 


44 


TRIGONOMETRY 


cot  (90°  +  «)=-  =  —  ^  =  -  tan  a 

y'     » 

sec  (90°  +  «)==  —  =  —  =  -  esc  « 
a;       —  y 

esc  (90°  +  «)=  —  =  -  =  sec  a 
' 


47.  Functions  of  90°  —  a  in  terms  of  functions  of  a  (n  =  1). 
Let  XOP  be  any  positive  angle  and  XOP'  the  angle 
90°  -  a. 

Then  taking  OP'  =  OP  we  have,  for  each  figure, 


x  = 


y 


x,  rf  =  r. 


Y 

Y 

D' 

7 

y     P 

&& 

•>.                   .•    v 

T> 

.X'                  0 

x'       x      x                   X'    J 

aj^V 

/                           Y 

y\/< 

^ 

•$* 

p'  ry 

/ 

Y 

\ 

/ 

D             ' 

Y 

Similar  figures  can  be  constructed  for  the  other  quadrants. 
Then  in  each  quadrant 

sin  (90°  -  a)  =  £  =  -  =  cos  a 

cos  (90°  -«)  =  -  =  %  =  sin  a 
r'      r 

tan(90°-<x)  =  ^  =  - 
x'     y 

cot  (90°  -  a)  =  -  =  ^  =  tan  a 

y     % 

sec  (90°  -  a)  =  —  =  -  =  esc  a 
x'      y 

esc  (90°  —  a)  =  -  =  -  =  sec  a 

y1    ^ 


VARIATIONS  OF  TRIGONOMETRIC   FUNCTIONS     45 


48.  Functions  of  180°  —  a  in  terms  of  functions  of  a(n=2). 
Let  XOP  be  any  positive  angle  and  XOP'  the  angle 
180°  -  a. 

Then  taking  OP'  =  OP  we  have,  for  each  figure, 

»'  =  -»,  y'=y,  r'  =  r. 


P'Y 

P 

Y 

«IY 

r 

\i 

Bg. 

2/ 

\ 

fij 

a?'  X5 

"^      a? 

X               •**  vj 

«                x         X'       2/' 

^S-J^   x 

P 

p 

r 

| 

Y 

Y 

Similar  figures  can  be  constructed  for  the  other  quadrants. 
Then  in  each  quadrant 

sin  (180°  -  «)  =  ^  =  ^  =  sin  « 

T         T 

cos  (180°  -  a)  =  -,=  —  =  -  cosa 

tan (180°  -«)  =  ^  =  -^-  =  -  tana 

x'      —  x 

cot  (180°  -  «)=  -  =  -^^  =  -  cot  a 
2/'        y 

sec  (180°  -«)  =  -  =  -^-  =  -  sec  a 
x'      —  x 

esc  (180°  -  a)  =  -  =  -=  csca 

y1    y 

49.  Laws  of  reduction.  The  method  of  the  last  four 
articles  may  be  applied  to  any  angle  of  the  form  n  90°  ±  a 
to  obtain  formulas  of  reduction  for  all  positive  and  negative 
integral  values  of  w,  a  being  any  angle.  '  A  complete  inves- 
tigation would  show  the  following  laws  to  be  true : 


I 


46  TRIGONOMETRY 

When  n  is  even,  any  function  of  n  90°  ±  a  is  numerically 
equal  to  the  same  function  of  a.  ;  when  n  is  odd,  any  function 
ofn  90°  ±  a  is  numerically  equal  to  the  cofunction  of  a. 

If  the  function  of  n  90°  ±  a  is  positive,  a  being  considered 
acute,  the  members  of  the  equation  have  like  signs  ;  if  the 
function  of  n  90°  ±  a.  is  negative,  a  being  considered  acute, 
the  members  of  the  equation  have  unlike  signs.  The  results 
thus  obtained  are  valid  for  all  positive  and  negative  values 
of  a. 

50.  EXAMPLES 

Express  as  functions  of  a  positive  angle  less  than  90°  : 

1.  sin  130°. 

SUGGESTION.     130°  =  90°  +  40°,  or  130°  =  180°  -  60°. 

2.  cos  170°.  5.   cos  (-20°). 

3.  tan  110°.  6.   tan  (-80°). 

4.  cot  160°.  7.   sin  (-120°). 

Express  as  functions  of  0 : 

8.  sin  (810° -0).  12.   tan  (0  -  180°). 

9.  tan(360°-0).  13.    sec  (-  180°  -  6). 

10.  cot  (270°  +  0).  14.   esc  (-  630°  +  0). 

11.  sin  (9  -  90°).  15.   cos  (990°  -  0). 

Express  each  of  the  trigonometric  functions  of  the  follow- 
ing angles  as  functions  of  a  without  using  the  laws  of 
reduction. 

16.  180°  +  a.  18.   270°  +  a. 

17.  270°  -a.  19.   360°  -a. 


CHAPTER   V 
FUNDAMENTAL  RELATIONS.    LINE  VALUES 

51.  In  the  present  chapter  eight  fundamental  relations 
between  the  trigonometric  functions  are  developed.  It  is 
then  shown  that  each  trigonometric  function  may  be  repre- 
sented geometrically  by  a  single  line,  giving  the  so-called 
line  values.  This  gives  a  second  view  of  the  trigonometric 
functions.  The  line  values  offer  a  simple  method  of  demon- 
strating the  fundamental  relations,  and  of  developing  the 
properties  already  derived  by  means  of  the  trigonometric 
ratios.  In  fact  the  line  values  might  serve  as  the  funda- 
mental definitions  of  the  trigonometric  functions  and  thus 
trigonometry  could  be  based  upon  these  values.  Inciden- 
tally the  line  values  suggest  the  origin  of  some  of  the  terms 
used  to  designate  the  several  trigonometric  functions. 

. FUNDAMENTAL  RELATIONS 


52.  Certain  relations  of  fundamental  importance  exist 
between  the  trigonometric  functions  of  an  angle.  These 
will  now  be  developed. 

From  definitions, 

sin  a  =  -, 
r 

csca=-.        .-.  sina  =  — - — .  (1) 

y  csc  a 


x 
cos  a  =  -. 


seca=-»       .-.  cosa=— - —  (2) 

x  sec  a 

47 


48  TRIGONOMETRY 


tan  a  =  ^ , 
x 

cot  a  = -.        .-.  tan  a  = 


cot  a 


sin  a  =  -j 


cos  a 


tan  a  =    .        .-.  =  tan  a.  (4) 

x  cos  a 


cota=-,      and      -        cot  a.  (5) 

sin  a 


From  geometry,  2/2  +  x2  =  r2. 

Dividing  successively  by  r2,  #2,  and  yz, 

2?  + 12  =  1,      .-.  sin2  a  +  cos2  a  =  1.  (6) 

•£  +  1=^,     ...  tan2  a  +  1  =  sec2  a.  (7) 

a2  x2 

^  +  1=^,     /.  cot2  a  +  1  =  esc2  a.  (8) 

These  eight  formulas  are  called  the  fundamental  relations. 
They  are  used  frequently,  and  should  be  memorized.  In 
doing  this,  it  is  well  to  have  the  method  of  derivation 
clearly  in  mind. 

53.  The  use  of  exponents.  In  affecting  trigonometric 
functions  with  exponents,  the  exponent  is  usually  placed 
immediately  after  the  function.  Thus,  in  the  formulas  of 
the  preceding  article,  sin2  a  is  identical  in  meaning  with 
(sin  a)2,  and  tan2  a  is  identical'  with  (tan  a)2. 

An  exception  to  the  above  is  made  for  the  exponent  —  1, 
in  which  case  it  is  always  necessary  to  make  use  of  paren- 


FUNDAMENTAL  RELATIONS  49 

theses.     Thus,   (cos  a)"1   is   used   to   express    ,   while 

cos  a 

cos"1  a  has  an  entirely  different  meaning,  as  will  be  ex- 
plained later. 

54.  Trigonometric   identities.     An  identical  equation,  or 
an  identity,  is  an  equation  which  is  satisfied  for  all  values 
of  the  unknown  quantities.     The  eight  fundamental  rela- 
tions are  trigonometric  identities.     There  are  many  other 
identities  depending  upon  these. 

The  truth  of  an  identity  can  be  established  in  two  ways : 

First.  Begin  with  one  of  the  fundamental  relations,  and 
produce  the  given  identity  by  means  of  the  fundamental 
relations  and  algebraic  principles. 

Second.  Begin  with  the  given  identity  and  transform  it 
to  one  of  the  fundamental  relations,  or  reduce  one  member 
of  the  equation  to  the  other  by  means  of  the  fundamental 
relations  and  algebraic  principles. 

The  choice  of  the  trigonometric  transformations  neces- 
sary to  effect  the  reductions  is  often  suggested  by  the 
functions  involved  in  the  given  problem.  When  no  trans- 
formation is  thus  suggested,  the  problem  may  generally  be 
simplified  by  expressing  each  of  the  functions  in  terms 
of  sines  and  cosines,  and  making  use  of  the  relation 
sin2  a  4-  cos2  a  =  1. 

Avoid  the  use  of  radicals  whenever  possible. 

55.  Trigonometric   equations.     A  conditional  equation  is 
an   equation  which   is  not   satisfied  for  all  values  of  the 
unknown  quantity,  but  is  satisfied  only  for  particular  values. 

(a)  Trigonometric  equations  involving  different  trigono- 
metric functions  of  the  same  angle  may  often  be  solved  by 
simplifying  the  equation  by  the  use  of  the  fundamental 
relations.  Thus  sin  a  _  COs  a 

may  be  written  in  a  =  1  or  tan  a  =  1 

cos  a 

/.a  =  45°  or  225°. 


50  TRIGONOMETRY 

(5)  Some  equations  may  conveniently  be  solved  by  trans- 
posing all  of  the  terms  to  the  first  member  of  the  equation, 
factoring,  and  then  placing  each  factor  equal  to  zero.  Thus 

2  sin2  a  +  V3  cos  a  =  2  V3  sin  a  cos  a  -f  sin  a 
may  be  written 

sin  a  (2  sin  a  —  1)  -f-  V3  cos  a  (1  —  2  sin  a)  =  0 
or 

(sin  a  —  V3  cos  a)  (2  sin  a  —  1)  =  0. 

This  equation  is  satisfied  if  either 

sin  a  —  V3  cos  a  =  0  or  2  sin  a  —  1  =  0 
whence  tan«=V3  or  sina  =  i. 

/.a  =  60°,  240°  or  a  =30°,  150°. 

(c)  When  no  other  method  suggests  itself,  the  equation 
may  be  transformed,  by  the  use  of  the  fundamental  rela- 
tions, into  an  equation  containing  only  one  function.  The 
equation  thus  obtained  may  be  solved  algebraically  for  the 
function  involved,  from  which  the  values  of  the  angle  may 

be  obtained.     Thus 

2 

10  cos3  a  tan  a  —  9  cos2  a 10  sin  a  +  9  =  0 

esc  a 

10  sin  «(1  -  sin2  a)-  9(1  -  sin2  a)—  12  sin  «  +  9  =  0 
10  sin  a  —  10  sin3  a  +  9  sin2  a  —  12  sin  a  =  0 
.-.  sin  a  =  0  or  10  sin2  a  —  9  sin  a  -f  2  =  0 
from  which  sin  a  =  0,  f ,  or  •£•. 

.-.  a  =  0°,  23°  35',  30°,  150°,  156°  25',  or  180°. 
A  trigonometric  equation  is  considered  completely  solved 
when  every  positive  angle  less  than  360°  which  satisfies  it 
has  been   determined.     All  other   angles   coterminal  with 
these  angles  also  satisfy  the  equation. 

In  the  solution  of  trigonometric  equations  extraneous 
roots  may  occur.  These  may  be  detected  by  substitution  in 
the  original  equation. 


FUNDAMENTAL  RELATIONS  51 

56.  EXAMPLES 

Prove  the  following  trigonometric  identities  : 

1.  sin  6  =  cos  0  tan  0.  cot  A 

3.   cos  6  = 

2.  tan  0  =  sin  0  sec  0.  esc  0 

4.  (1  —  sin  #)(!  +  sin  x)  =  cos2  x. 

5.  (sec  x  —  tan  #)(sec  x  +  tan  a?)  =  1. 

6.  (sin  x  -f-  cos  #)2  =  2  sin  a;  cos  cc  -}-  1. 


7. 


cot  a; 

8.  cos2  a  tan2  «  +  sin2  a.  cot2  «  =  1. 

9.  Given  cos  a.  =  -|  ,  a  being  in  the  fourth  quadrant,  find 
the  values  of  the1  remaining  trigonometric  functions. 

SOLUTION.     Since  sin2  a  +  cos2  a  =  1, 

sin2  a  =  1  -  cos2  a  =1  -  |f  =  |f. 
.*.  sin  a  =  —  \/|f  =  —  $\/l4. 
sin  a 


Also  tan  a  = 

cot  a  = 


esc  a  = 

sin  a      —  fv/14         2\/14 

Compare  with  the  method  of  examples  26  and  27,  Art.  27. 

10.    Given  tan  y  —  4,  y  being  in  the  third  quadrant,  find 
the  values  of  the  remaining  trigonometric  functions. 
SOLUTION.     Since  sec2  y  =  I  -f  tan2  y, 

sec  y  =  —  Vl  -f  16  =  —  \/17. 

Also  cot  y  = =  -  ; 

tan  y     4 

esc2  y  =  1  +  cot2  y  =  1  +  ^  =  £f ; 
esc  y  =  — 


52      I/  TRIGONOMETRY 

sin  3,  =  -!-  =  — 1—  =  -AVl7; 


cos  y  =  —  =  —  I—  =  -  TV  VI7. 
secy     _vi7 

11.  Given  sin  0  =  -|,  0  terminating  in  the  second  quadrant, 
find  the  values  of  the  remaining  trigonometric  functions. 

12.  Given  cot  0  =  T7^,  0  terminating  in  the  first  quadrant, 
find  the  values  of  the  remaining  trigonometric  functions. 

13.  Given   sec  0  =  —  2,  0  being   in  the   third  quadrant, 
find  the  values  of   the  remaining  trigonometric  functions. 

14.  Given  tan  a  =  —  -£,  find  the   remaining  functions  of- 
the  angles  less  than  360°  which  satisfy  the  equation. 

Solve  the  following  trigonometric  equations  for  angles 
less  than  90°: 

15.  5  sin  x  +  8  =  3(4  —  sin  x). 

16.  sinw  —  cos  w  =  0. 

17.  2  cos2  x  —  3  cos  x  4-  1  =  0. 

18.  (tan  6  -  l)(tan  0  -  V3)  =  0. 

19.  Solve  sin  x  =  cot  x  for  sin  x. 

Some  of  the  following  examples  are  identities  and  others 
are  equations.  Establish  the  identities  and  solve  the  equa- 
tions for  angles  less  than  180°. 

20.  sin4  x  =  1  —  2  cos2  x  +  cos4  x. 

21.  sin4  x  =  2  —  6  cos2  x  +  cos4  x. 


22.    (V§  +  l)L+  =  sin 

tan  6 


23    2Bin!  +  costf  VS 

tan  $     cot  $  2 

24.  tan  u  +  cot  u  =  sec  u  esc  u. 

25.  2  sin2  y  esc  ?/  +  3  esc  y  =  1. 


FUNDAMENTAL  RELATIONS  53 

Prove  the  following  trigonometric  identities  : 

26.  tan*  0  +  cot2  0  =  sec2  0  esc2  6-2.    . 

27.  sec2  ft  +  cos2  (3  =  tan2  ft  sin2  ft-}-  2. 

28.  esc2  y  —  sec2  y  =  cos2  y  esc2  y  —  sin2  y  sec2  y. 

rtrt       .  sin  a  4-  tan  a 

29.  sin  a  tan  ce  =  —  —  • 

cot  «  +  csc  <* 

30.  cot2  a  —  cos2  a  =  cot2  a  cos2  a. 


31. 


sin  0  cos  0 

32.  (sin  0  4-  cos  0)2  4-  (sin  0  -  cos  0)2  =  2. 

33.  sin2  x  4-  vers2  a?  =  2(1  —  cos  x). 

34.  sec2  a  csc2  a  _(l~tan2a)2=  4 

tan2  a 

Solve  the   following  trigonometric  equations  for  angles 
less  than  180°: 

35.  2  cos2  a  —  3  sin  a  =  0. 

36.  sec2  a  4-  cot2  a  =  4±, 

37.  1  4-  tan2  a;  —  4  cos2  a;  =  0. 

38.  tan  x  4-  cot  x  =  2. 

39.  2  sin2  x  4-  3  cos  x  ==  0. 

40.  V3  cos  x  4-  sin  #  =  V2. 

41.  csc2  a:  —  4  sin2  a;  =  0. 

42.  cot  aj  4-  csc2  x  =  3. 

LINE  VALUES 

57.   Representation  of  the  trigonometric  functions  by  lines. 

The  trigonometric  functions  of  an  angle  have  been  studied 
solely  from  the  standpoint  of  ratios  ;  but  each  function  can 
also  be  represented  in  magnitude  and  sign  by  a  single  line. 


54 


TRIGONOMETRY 


Let  LOP  be  any  angle.  Take  OP=l.  Draw  the  ordi- 
nate  and  abscissa  of  the  point  P.  About  0  as  the  center 
construct  a  circle  with  OP  as  a  radius.  Draw  the  tangents 
to  this  circle  at  L  and  F,  the  beginning  of  the  first  and  second 
quadrants  respectively,  and  produce  them  to  intersect  the 
terminal  line,  or  the  terminal  line  produced,  in  M  and  G. 


Then 

y     AP     AP 

sin  «  =  -  =  — —  =  — —  • 


OP 


AP 


x     OA     OA      KA 

cos  a  =  -  = =  — —  =  OA 

r      OP       1 


x     OA      OL 
=  x=OA  =  FG==FG 


OP 


x     OA      OL 


FUNDAMENTAL   RELATIONS  55 


or 

Thus  the  trigonometric  functions  are  represented  by  the 
segments  AP,  OA,  LM,  FG,  OM  and  OG.  It  is  evident 
that  these  segments  represent  the  trigonometric  functions 
in  magnitude.  They  also  represent  the  trigonometric  func- 
tions in  sign.  In  accordance  with  the  conventions  of  Arts. 
6  and  11,  the  segments  AP  and  LM  are  positive  when  drawn 
upward  from  the  X-axis  and  negative  when  drawn  down- 
ward ;  the  segments  OA  and  FG  are  positive  when  drawn 
from  the  F-axis  to  the  right  and  negative  when  drawn  to 
the  left;  the  segments  OM  and  OG  are  positive  when  they 
coincide  with  the  terminal  line  of  the  angle  and  negative 
when  they  coincide  with  the  terminal  line  produced. 

It  is  evident  that  the  sign  of  each  trigonometric  function 
is  determined  by  the  segment  representing  the  function, 
since  in  each  quadrant  the  segment  is  positive  whenever  the 
ratio  defining  the  function  is  positive,  and  negative  whenever 
the  ratio  is  negative.  Thus  in  each  quadrant  LM  has  the 

same  sign  as  the  ratio  ?,  and  similarly  for  the  other  func- 
x 

tions.     Therefore  the  segments  represent  the  trigonometric 
functions  in  sign  as  well  as  in  magnitude. 

The  segments  which  represent  the  trigonometric  functions 
are  called  the  line  values  of  the  functions. 

PROBLEM.     Represent  vers  a  and  covers  a  by  line  values. 

58.  Variations  of  the  trigonometric  functions  as  shown  by 
line  values.  The  line  values  of  the  trigonometric  functions 
give  a  simple  method  of  tracing  the  variations  in  the  func- 
tions as  the  angle  varies  from  0°  to  360°.  Thus  as  the  point 
P,  in  the  figures  of  Art.  57,  describes  the  circle  of  radius 
unity,  the  changes  in  the  segments  AP,  OA,  LM,  etc., 
represent  the  variations  in  the  sine,  cosine,  tangent,  etc., 
respectively. 


56 


TRIGONOMETRY 


Numerous  figures,  with  three  or  four  values  of  the  angle  in 
each  quadrant,  serve  to  suggest  these  variations. 

59.  Fundamental  relations  by  line  values.  By  the  use  of 
line  values  the  fundamental  relations  of  Art.  52  may  be 
simply  obtained  and  easily  memorized. 


F« cot  a -, 


Thus 

From  the  triangle  OAP, 

AP2  +OA2=OP2        or        sin2  a  +  cos2a  =  1. 
From  the  triangle  OLM, 

or        tan2  a  + 1  =  sec2  a. 
or        cot2a+l  =  csc2a. 


From  the  triangle  OGF, 


From  definitions, 

AP     sin  a 

tan  a  = = 

OA     cos  a 

OA     cos  a 


cot  a 


AP     sin  a 

sec«  =  -— = 

OA     cos  a 

esca-O?.-1 


AP     sin  a 


or 
or 
or 
or 


tana  = 
cot  a  = 
seca  = 
csca  = 


sin  a 
cos  a 
cos  a 
sin  a 


cos  a 

1 
sin  a 


FUNDAMENTAL  RELATIONS  57 


co«_=        =  __    or 

AP     LM     tan  a  tan  a 

The  process  of  derivation  of  these  formulas  applies 
equally  well  to  an  angle  of  the  third  or  fourth  quadrant  ; 
hence  the  formulas  are  true  for  all  values  of  a. 

To  memorize  these  formulas  most  easily,  form  a  clear 
mental  picture  of  the  figure  for  the  first  quadrant  and  read 
each  formula  directly  from  the  figure,  applying  the  ratio 
definitions  and  the  Pythagorean  theorem. 

60.  EXAMPLES 

1.  Construct  the  line  values  of  an  angle  terminating  in 
the  third  quadrant. 

2.  Construct  the  line  values  of  an  angle  terminating  in 
the  fourth  quadrant. 

3.  Obtain  the  fundamental  relations  for  an  angle  in  the 
third  quadrant,  by  the  use  of  line  values. 

4.  Obtain  the  fundamental  relations  for  an  angle  termi- 
nating in  the  fourth  quadrant,  by  the  use  of  line  values. 

5.  Deduce  the  relations  between  the  functions  of  90°  +  x 
and  the  functions  of  x  by  means  of  line  values. 

SUGGESTION.  Construct  two  figures,  one  for  the  angle  90°  +  x  and 
the  other  for  the  angle  x. 

6.  Deduce  the  relations  between  the  functions  of  180°  —  x 
and  the  functions  of  x  by  means  of  line  values. 

7.  Deduce  the  relations  between  the  functions  of  90°  —  x 
and  the  functions  of  270°  +  x  by  means  of  line  values. 

By  means  of  line  values,  trace  the  variations  in  the  follow- 
ing functions  as  a  varies  from  0°  to  360°  : 

8.  sin  a.  11.    cot  «. 

9.  cos  «.  12.    sec  a. 
10.   tan  a.                      13.   esc  a. 


58  TRIGONOMETRY 

Prove  the  following  trigonometric  identities  : 

14.    =  sin  x  cos  x. 

cot  x  +  tan  x 

15.  cos2  a  (1  +  cot2  a)  =  cot2  a. 

16.  sin  0  cot2  0  =  esc  0  -  sin  0. 

17.  tan  a  —  tan  «  sin2  a  =  sin  a  cos  a. 

18.  cos  a  tan2  a  =  sec  a  —  cos  a. 

jo     cos  «      sin  «  _  sin3  a  -f-  cos3  a 
tan  «     cot  a         sin  «  cos  a 

on    sin3  a  +  cos3  «      .< 

20.  — =  1  —  sin  a  cos  a. 

sin  a  ,  cos  a 

Oi     sin  x  .  tan  x  .  sec  x     2  cot  a?  4- 1 

1 


22. 

23. 

04 


cosa;      cotx      csca;          cot2  re 

1  —  cos  x  _  sec  a;  —  1 
1  -|-  cos  a?      sec  x  +  1 

1  —  cos  x  sin2  a; 


1  +•  cos  x      (1  +  cos  a:) 
1  —  cos  #      (1  —  cos  #2 


1  -j-  cos  a?  sin'"  a; 

25.  2  sin  ?/  cos2  y  +  (2  cos2  ?/  —  1)  sin  y  =  3  sin  y  —  4  sin3  y. 

26.  cos4  a  —  sin4  a  =  cos2  a  —  sin2  a. 

27.  sec2  a  +  cosec2  a  =  sec2  a  esc2  a. 

no    tan  a;  -f-  tan  y  _  sec  x  -+-  sec  y 
\         sec  a;  —  sec  ?/      tan  a;  —  tan  y 

29.  tan4  w  —  cot4  u  =  (tan2  u  +  cot2  w)(sec2  w  —  esc2  w). 

on    tan  a;  -f  tan  ?/ 

30.  -    —  -i-  -  ?  =  tan  x  tan  y. 

cot  x  +  cot  y 

01     tana?  —  tan  y 

31.  -    -  2  ==  —  tan  x  tan  w. 
cot  x  —  cot  ?/ 


FUNDAMENTAL   RELATIONS  59 

on    tan  x  -f  cot  y     tan  x 
64. =  —    —  • 

cot  x  -f-  tan  y   •  tan  y 

33.  Express  sin2  a;  —  cos2  a;  in  terms  of  tan  x. 

34.  Express  vers  x  —  covers  x  in  terms  of  sin  x. 

35.  Express  cot2  a?  -f-  csc2#  in  terms  of  cos  x. 

•yp-pQ    /V» 

36.  Express  —   —  in  terms  of  tan  x. 

covers  x 

37.  Express  tan2  x  +  sec2  x  in  terms  of  cot  x. 

38.  Express  '(4  sin3  0  -  3  sin  0)(2  cos2  0-1) 

-f-  (3  cos  0  —  4  cos3  6)  (2  sin  6  cos  0)   in 
terms  of  sin  0. 

39.  Given  sin  0  =  a,  find  the  remaining  functions  of  0. 

40.  Given  cot  0  =  6,  find  the  remaining  functions  of  0. 

41.  Express  each  trigonometric  function  of  0  in  terms  of 
cos  0. 

42.  Express  each  trigonometric  function  of  0  in  terms  of 
tan0. 

43.  Simplify  (a  +  6)  sin  30P  -  6  cos  60°  +  a  tan  180°. 

44.  Simplify  I  sin  (270°  -  a;)  +  m  cos  (180°  +  a?) 

+  w  sin  (90°  -  a;). 

45.  Simplify  a  sin  135°  +(a  —  6)  cos  225°  +  &  cos  315°. 

46.  Simplify  tan  (-120°)  +  cot  150°-tan  210°+  cot  240°. 

Find  the  positive  values  of  0,  less  than  180°,  that  satisfy 
the  following  equations : 

47.  sin0  cos  0  =  0. 

48.  sin  0  +  cos  0  =  0. 

49.  sin0(2sin0-l)(2cos0-l)  =  0. 

50.  sin  0  +  cos  0  cot  0  =  2. 


60  TRIGONOMETRY 

51.  sin0cos0  =  |. 

52.  2  sin  0  cos  <9  + sin  0-2  cos0  =  l. 

53.  2  cos  0  +  sec  6  =  3. 

54.  sec  (9  +  tan  0  =  2. 

55.  2  sin  0  +  5  cos  9  =  2. 

56.  1  +  sin2  0  =  3  sin  0  cos  0. 

57.  tan4  0-4  tan2  0-5  =  0. 

58.  What  negative  angles,  numerically  less  than  180°, 
satisfy  the  equation  sin2  x  -f-  sin  x  =  cos2  x  —  1  ? 

59.  Given  9  cos2  w  +  9  sin  ^  =  11,  find  tan  u. 

60.  Given  tan2  x  —  5  sec  a;  +  7  =  0,  find  sin  x. 

Find  the  positive  values  of  0,  less  than  360°,  that  satisfy 
the  following  equations : 

61.  sin  0  =  - cos  285°. 

62.  tan  0  =  cot  (-144°). 

63.  cos  ( -  0)  =  sin  190°. 

64.  sin0  =  —  sin  50°. 

65.  If  sin  122°  =  k,  prove  that  tan  32°  =  ~V/J 

66.  If  cot  255°  =  a,  prove  that  cos  345°  = 

67.  If  cos(-  100°)=  k,  prove  that  tan  80°  =  - 

k 

Solve  the  following  equations  for  any  trigonometric  func- 
tion of  a. 

68.  2  sin  a  +  esc  a  =  1.  71.   tan2  a  -f  4  sec  a  =  5. 

69.  esc  a  cot  a  =  j.  72.   tan  a  +  cot  a  =  2. 

70.  2  sin  a  +  cos2  a  =  1. 

73.    Given  sin  ^  =  k  sin  v  and  tan  u  =  Z  tan  v,  find  sin  w 
and  sinv. 


CHAPTER   VI 


FUNCTIONS   OF   THE    SUM   OF  TWO   ANGLES 
DOUBLE  ANGLES.     HALF  ANGLES 

61.  We  have  thus  far  considered  the  properties  and  rela- 
tions of  the  functions  of  a  single  angle.  We  have  also 
shown  that  the  functions  of  the  angles  n  90  ±  a  depend 
upon  the  functions  of  the  angle  a. 

In  the  present  chapter  we  consider  the  relations  between 
the  functions  of  the  sum  of  two  independent  angles  and  the 
functions  of  the  separate  angles,  and  develops  some  related 
formulas. 

-*  62.   The  sine  of  the  sum  of  two  acute  angles  expressed  in 
terms  of  the  sines  and  cosines  of  the  angles. 


B     X 


Let  a  and  /?  be  any  given  acute  angles. 

Construct  the  angle  (a  -f-  /?),  Art.  12. 

Then  (a  +  ft}  may  be  acute  as  in  Fig.  1,  or  obtuse  as  in 
Fig.  2. 

From  any  point  P,  in  the  terminal  line  of  the  angle 
(a  4-  /?),  draw  PA  perpendicular  to  OX,  and  PQ  perpendicu- 
lar to  OS.  Through  Q  draw  RQT  parallel  to  OX,  and  QB 
perpendicular  to  OX. 

61 


62  TRIGONOMETRY 

Then  the  angle  TQP  is  equal  to  90°  +  a. 
By  definition 


OP          OP          OP     OP 

BQ.OQ.RP   QP 
OQ'  OP    QP'  OP 


and         =s 


therefore  sin  (a  +  P)  =  sin  a  cos  p  4-  cos  a  sin  p.  (1) 

PROBLEM  1.     Show  that  formula  (1)  is  true  when  either 

angle  is  90°. 

PROBLEM  2.     Show  that  formula  (1)  is  true  when  each 

angle  is  90°. 

63.  The  cosine  of  the  sum  of  two  acute  angles  expressed  in 
terms  of  the  sines  and  cosines  of  the  angles.  '  Referring  to  the 
figures  of  the  previous  article,  we  have  by  definition 


OB  OQ    QR  QP 
OQ'  OP    QP'  OP 


But        =  cos  «,         =  cos  ft 


=cos  (90°  +«)  =  -  sin  a,  and         =  sinft 

therefore  cos  (a  -f  p) =cos  a  cos  p  —  sin  a  sin  p.  *"  (1) 

PROBLEM  1.     Show  that  the  formula  (1)  is  true  when 

either  angle  is  90°. 

PROBLEM  2.     Show  that  formula  (1)  is  true  when  each 

angle  is  90°. 

64.   The  formulas  developed  in  the  last  two  articles  are 
of  the  utmost  importance,  since  many  other  formulas  are 


FUNCTIONS  OF  THE  SUM  OF  TWO  ANGLES      63 

derived  from  them.  They  form  the  basis  of  the  present 
chapter.  In  developing  these  formulas  a  and  ft  were  con- 
sidered positive  acute  angles,  but  the  formulas  are  true  for 
all  values  of  a  and  ft}  as  may  be  shown  either  geometrically 
or  analytically. 

It  is  shown  geometrically  by  constructing  figures  in 
which  a  and  (3  are  of  any  magnitude,  and  following  the 
method  of  proof  given  above  for  acute  angles. 

We  present  the  analytic  demonstration  as  the  more  satis- 
factory. 

65.    To  prove  that 

sin  («  -|-  ft)  =  sin  a  cos  ft  4-  cos  a  sin  ft  (1) 

and  cos  (a  -f-  ft)  =  cos  a  cos  ft  —  sin  a  sin  ft  (2) 

are  true  for  all  values  of  a  and  ft. 

First.  To  show  that  a  can  be  replaced  by  a',  where  a'  = 
a  +  90°. 

Let  a'  =  a  -f  90°  where  a  is  acute. 

As  a  varies  from  0°  to  90°,  a'  varies  from  90°  to  180°. 

Also  sin  a'  =  sin  (a  +  90°)  =  cos  a,  (3) 

cos  a'  =  cos  (a  +  90°)  =  —  sin  a.  (4) 

Now  sin  («'  +  /?)=  sin  (90°  +  a  +  £)=  cos(a  -f  £). 

Since  a  and  /?  are  both  acute,  cos  (a  +  ft)  may  be  ex- 
panded by  (2),  hence 

sin  (a'  +  ft)  =  cos  (a  +  ft)  =  cos  a  cos  ft  —  sin  a  sin  ft. 
Therefore,  by  (3)  and  (4), 

sin  (a'  +  ft)  =  sin  a'  cos  ft  +  cos  «'  sin  ft.  (5) 

Similarly, 

cos(«'  -f  ft)  =  cos  (90°  +  a  +  ft)  =  -  sin  («  -f  /8) 
=  —  sin  a  cos  /?  —  cos  a  sin  /?. 


64  TRIGONOMETRY 

Therefore,  by  (3)  and  (4), 
cos  («'  +  /?)  =  cos  a'  cos  ft  —  sin  a'  sin  j3.  (6) 

Formulas  (5)  and  (6)  show  that  one  angle,  a,  in  (1)  and 
(2)  may  be  extended  to  180°. 

Second.  To  show  that  a  can  be  replaced  by  a"  where  a"  = 
a  +  18CP- 

Let  a"  =  a  +  180°. 

As  a  varies  from  0°  to  180°,  a!'  varies  from  180°  to  360°. 
Also  sin  a"  =  sin  (a  +  180°)  =  —  sin  a,  (7) 

cos  a"  =  cos  (a  +  180°)=  —  cos  a.  (8) 

Now  sin  (a"  -f  0)=  sin  (180°  +  a  +  /?)=  -  sin  (a  +  0). 

Since  a  is  less  than  180°  and  ft  acute,  sin  («  4-  /?)  may  be 
expanded  by  (1),  as  extended  in  the  first  part  of  the  proof, 
hence 

sin  (a"  +  /?)  =  —  sin  (a  +  ft)  =  —  sin  a  cos  ft  —  cos  a  sin  ft. 

Therefore,  by  (7)  and  (8), 

sin  (a"  +  ft)  =  sin  a!'  cos  ft  -f  cos  a"  sin  ft.  (9) 

Similarly, 

cos  (a"  4-  /?)=  cos  (180°  4-  a  +  0)  =  -  cos  (a  +  0) 

=  —  cos  a  cos  /?  +  sin  a  sin  /3. 
Therefore,  by  (7)  and  (8), 

cos  (a"  -f  ft}  =  cos  a"  cos  0  —  sin  a"  sin  /?.  (10) 

Formulas  (9)  and  (10)  show  that  one  angle,  a,  of  (1)  and 
(2)  may  be  extended  to  360°. 

Third.  In  a  similar  manner  it  may  be  shown  that  (1)  and 
(2)  are  true  when  ft  varies  from  0°  to  360°,  a  having  any 
value  from  0°  to  360°,  from  which  it  easily  follows  that  (1) 
and  (2)  are  true  for  all  positive  values  of  a  and  ft. 


FUNCTIONS   OF  THE   SUM  OF  TWO  ANGLES      65 

Fourth.  Formulas  (1)  and  (2)  may  be  shown  to  be  true 
when  either  a  or  ft  or  both  are  negative,  by  the  use  of  the 
substitutions 

a'  =  a-n  360° 


hence  (1)  and  (2)  are  true   for  all   positive  and  negative 
values  of  a  and  (3. 

66.    To  find  the  tangent  of  the  sum  of  any  two  given  angles 
in  terms  of  the  tangents  of  the  given  angles. 
Let  a  and  ft  be  the  given  angles. 

Then  tanQ  +  8)  =  sin  ("  +  ffl  =  sina  cos  ft  +  cos  «  sin  ^ 
cos  (a  +  ft)      cos  a  cos  ft  —  sin  a  sin  ft 

Dividing  numerator  and  denominator  of  the  last  fraction 
by  cos  a  cos  ft  and  simplifying,  we  have 


1  —  tan  a  tan  p 

67.    To  find  the  cotangent  of  the  sum  of  any  two  given 
angles  in  terms  of  the  cotangents  of  the  given  angles. 
Let  a  and  ft  be  the  given  angles. 

Then  cot  («  +  /?)  =  cos  <«  +  ^  =  cos  «  cos  /3  -  sin  «  sin/?. 

sm  (a  -h  ft)      sm  a  cos  ft  +  cos  a  sm  ft 

Dividing  numerator  and  denominator  of  the  last  fraction 
by  sin  a  sin  ft  and  simplifying  we  have 


cot  p+  cot  a 

68.   Addition  formulas.     Collecting  the  formulas  for  the 
sum  of  two  angles  for  convenience  of  reference,  we  have 

sin  (a  -h  ft)  =  sin  a  cos  ft  +  cos  a  sin  ft  (1) 

cos  (a  +  ft)  =  cos  a  cos  ft  —  sin  a  sin  /3  (2) 


66  TRIGONOMETRY 


69.    To  find  the  sine,  cosine,  tangent,  and  cotangent  of  the 
difference  of  two  given  angles. 

Since  the  formulas  of  Art.  68  are  true  for  all  values  of  ot 
and  ft  they  are  true  when  —  ft  is  substituted  for  ft. 

Hence 

sin  (a  —  ft)  =  sin  a  cos  (—  ft)  +  cos  a  sin  (—  ft) 
cos  (a  —  ft)=  cos  a  cos  (—  ft)  —  sin  a  sin  (—  ft) 

tan  (ct      K\  -  tang  + 
~^- 


cot(-0)+cota 
But  remembering  that 

sin  (—  ft)  =  —  sin  ft,  cos  (—  ft)  =  cos  ft 
tan  (-  ft)  =  -  tan  ft,  and  cot  (-£)  =  -  cot  ft 
these  formulas  become  U 

sin  (a  —  ft)  =  sin  a  cos  ft  —  cos  «  sin  /?  (1)  Ir. 

cos  (a—  ft)=  cos  a  cos  £  +  sin  a  sin  ft  (2) 

tan  (a  -  ft)  =    an<*~ — ?L/L  (3) 

1  4-  tan 


'  cot/3-cot« 

70.  EXERCISES 

1.   Find  sin  75°. 

SOLUTION.     75°  =  sin  (45°  +  30°)  =  sin  46°  cos  30°  +  cos  45°  sin  30° 


FUNCTIONS  OF  THE  SUM  OF  TWO  ANGLES   67 

2.  Find  cos  75°.  7.  Find  sin  (90°  -  £). 

3.  Find  sin  15°.  8.  Find  tan  (90°  +  a). 

4.  Find  cos  15°.  9.  Find  cos  (180°  -  a). 

5.  Find  tan  75°.  10.  Find  sec  15°. 

6.  Find  cot  15°.  11.  Find  esc  (90°  +  a). 

71.  Double  angles.  To  find  the  sine,  cosine,  tangent,  and 
cotangent  of  twice  a  given  angle  in  terms  of  the  functions  of 
the  given  angle. 

From  the  formulas  of  Art.  68,  letting  /3  =  a,  we  have, 
after  reduction, 

sin  2  a  =  2  sin  a  cos  a  (1) 

cos  2  a .  =  cos2  a  —  sin2  a  (2) 

=  l-2sin2a  (3) 

=  2  cos2  a  -  1  (4) 

rt  2  tan  a  /K-, 

tan  2  a  =  —  (5) 

1  -  tan2  a 

cot2a  =  cot2a-1.  (6) 

2  cot  a 

To  express  clearly  the  real  significance  of  these  formulas, 
we  may  state  them  from  two  points  of  view. 

If  a  be  the  angle  under  consideration,  the  formula 

sin  2  a  =  2  sin  a  cos  a 

may  be  stated :    The  sine  of  twice  an  angle  is  equal  to  twice 
the  sine  of  the  angle  times  the  cosine  of  the  angle. 

If  2  a  be  the  angle  under  consideration,  the  same  formula 
may  be  stated :  TJie  sine  of  an  angle  is  equal  to  twice  the 
sine  of  half  the  angle  times  the  cosine  of  half  the  angle. 

It  then  follows  that 

sin  a  =  2  sin  J  a  cos  |  a.  (7) 

Similarly,  if  a  be  the.  angle  under  consideration,  the 
formula  cos  2  a.  =  cos2  a  —  sin2  a 


68  TRIGONOMETRY 

may  be  stated  :  The  cosine  of  twice  an  angle  is  equal  to  the 
square  of  the  cosine  of  the  angle  minus  the  square  of  the  sine 
of  the  angle. 

If  2  a  be  the  angle  under  consideration,  the  same  formula 
may  be  stated  :  The  cosine  of  an  angle  is  equal  to  the  square 
of  the  cosine  of  half  the  angle  minus  the  square  of  the  sine  of 
half  the  angle. 

It  then  follows  that 

cos  a  =  cos2  1  a  —  sin2  ^  a.  (8) 

Similarly, 
from  cos  2  a  =  1  —  2  sin2  «  we  have 

cosa  =  l  -2sin2ia,  (9) 

from  cos  2  a  =  2  cos2  a  —  1  we  have 

l,  (10) 


from  tan  2  a  =    2  tan  "     we  have 
1  —  tan2  a 

2  tan  i  a 


from  cot  2  a  =  cot2  a  ~  l  we  have 
2  cot  a 

cota^****-1.  (12) 

2  cot  |a 

72.  Half  angles.  To  find  the  sine,  cosine,  tangent,  and  co- 
tangent of  one  half  a  given  angle  in  terms  of  functions  of  the 
given  angle. 

From  formula  (9),  Art.  71,  we  have 
cos  «=  1  —  2  sin2  ~j 


FUNCTIONS  OF  THE   SUM  OF  TWO  ANGLES      69 


From  formula  (10),  Art.  71,  we  have 

cos  a  =  2  cos2  -  —  1, 

2 


or  COS^  =  ±A/-        "~  (2) 

&  A 

From  (1)  and  (2), 

^^  (3) 


2          ^  1  +  cos  a 
and  cot 


2 

In  each,  of  these  formulas  the  positive  or  negative  sign 

is  chosen  to  agree  with  the  sign  of  the  function  of  ^  ,  depend- 

2i 

ing  on  the  quadrant  in  which  -  lies. 

2 

These  formulas  may  be  considered  from  two  view  points. 
For  example,  formula  (1)  may  be  stated  :  The  sine  of  half 
an  angle  is  equal  to  the  square  root  of  the  fraction  whose 
numerator  is  one  minus  the  cosine  of  the  angle,  and  whose 
denominator  is  two;  or,  Tlie  sine  of  an  angle  is  equal  to  the 
square  root  of  the  fraction  whose  numerator  is  one  minus  the 
cosine  of  twice  the  angle,  and  whose  denominator  is  two. 

73.  To  find  the  sum  and  difference  of  the  sines  of  any  two 
angles,  also  the  sum  and  difference  of  the  cosines  of  any  two 
angles. 

From  the  formulas  of  Arts.  68  and  69  we  have,  by  ad- 
dition and  subtraction, 

sin  (a  +  ft)  +  sin  (a  —  ft)  =  2  sin  a  cos  (3 
sin  (a  -f-  ft)  —  sin  (a  —  (3)  =  2  cos  a  sin  ft 
cos  (a  +  ft)  +  cos  (a  —  ft)  =  2  cos  a  cos  ft 
cos  (a  +  ft)  —  cos  (a  —  ft)  =  —  2  sin  a  sin  ft. 


70  TRIGONOMETRY 

Let  a  -J-  ft  =  y  and  a  —  ft  =•  d.  Solving  for  a  and  /?,  we 
have  a  =  £(y  4- 3)  and  ft  =  1  (y  —  d).  Substituting  these 
values  in  the  above  equation,  it  follows  that 

sin  Y  +  sin  d  =  2  sin  £  (y  +  d)  cos  1  (Y  -  3),  (1) 

sin  Y-  sin  3  =  2  cos  1  (y  +  3)  sin  |  (>-  3),  (2) 

cos  Y  4-  cos  3  =  2  cos  J  (Y  4-  3)  cos  J  (?  —  3),  (3) 

cos  Y  —  cos  d  =  —  2  sin  $(i  +  d)  sin  -J  (<y  —  3).  (4) 

Equations  (1),  (2),  (3),  and  (4)  may  be  read  from  two 
view  points. 

Eegarding  y  and  3  as  the  given  angles,  (1)  may  be 
stated :  The  sum  of  the  sines  of  two  angles  is  equal  to  twice 
the  product  of  the  sine  of  half  the  sum  of  the  given  angles  into 
the  cosine  of  half  the  difference  of  the  given  angles. 

Thus,  sin  6  x  -f-  sin  4  x  =  2  sin  5  x  cos  x. 

Eegarding  i(y  +  3)  and  -J(y  —  3)  as  the  given  angles,  it  is 
clear  that  their  sum  is  y  and  their  difference  3.  Then,  by 
reading  the  second  member  first,  equation  (1)  may  be  stated : 
Twice  the  sine  of  any  angle  times  the  cosine  of  any  other  angle 
is  equal  to  the  sine  of  the  sum  of  the  angles  plus  the  sine  of  the 
difference  of  the  angles. 

Thus,  2  sin  20°  cos  5°  =  sin  25°  +  sin  15°. 

74.  Equations  and  identities.  The  formulas  developed  in 
the  present  chapter  are  true  for  all  values  of  the  angles 
involved ;  hence  they  are  trigonometric  identities.  By  the 
use  of  these  identities  many  others  may  be  established. 
The  remarks  of  Art.  54,  concerning  the  use  of  the  funda- 
mental relations  in  establishing  identities,  apply  here. 

The  identities  of  the  present  chapter  are  also  useful  in 
solving  trigonometric  equations.  By  their  aid  an  equation 
involving  functions  of  multiple  angles  may  be  transformed 
into  an  equation  containing  functions  of  a  single  angle 
(see  Ex.  33,  Art.  75).  This  transformed  equation  can  then 
be  solved  as  indicated  in  Art.  55.  Frequently  equations 


FUNCTIONS   OF   THE   SUM  OF  TWO   ANGLES     71 

may  be  much  simplified  by  reducing  sums  or  differences  of 
sines  and  cosines  to  products  by  the  relations  of  Art.  73 
(see  Ex.  41,  Art.  75). 

75.  EXAMPLES 

1.  Find  sin  221°,  cos  221°,  tan  22J°,  and  cot  22£°. 

2.  Given    cosa  =  J;    find    sin  2  a,  cos  2  a,    tan2«,   and 
cot  2  a. 

3.  Given  tan  a  =  3 ;    find   sin  1  a,   cos  %  a,   tan  %  a,   and 
cot  \  a. 

Prove  the  following  identities : 

4.  (cos  a  +  sin  a)  (cos  a  —  sin  a)  =  cos  2  «. 

5.  (sin  a  +  cos  a)2  =  1  -f  sin  2  a. 

6.  sin  3  a  =  3  sin  a  —  4  sin3  a. 

SUGGESTION  :  sin  3  a  =  sin  (2  a  +  a). 

7.  cos3a  =  4cos3a  —  3 cos  a. 

8.  tan3«  =  3tan"-tan3". 

l-3tan2a 

1  +  COS  X 

,  f.                       sin  x 
10.   cot  i  x  = 

1  —  cos  x 

-  -     1  +  tan  i  x  _  1  4-sin^ 

1  —  tan  i  x  ~     cos  x 

-  ft  1  —  cos  2  a 

12.  tan2a=- 

1  +  cos  2  « 

13.  sec2«=     sec  tt  — 

2  — sec2« 

14.  4  sin2 1  a  cos2  i  a  =  1  —  cos2  «. 

15.  sin4«  +  sin2  «  =  2sin  3acos  a.     (See  Art.  73.) 

16.  cos  6  a  —  cos  2a=  —  2  sin  4  a  sin  2 a. 


72  TRIGONOMETRY 

17.  sin  (45°  +  «)  —  sin  (45°  -  a)  =  V2  sin  a. 

18.  sin  (30°  +  a)  +  sin  (30°  -  a)  =  cos  a. 

19.  Express  cos  4  a  sin  3  a  as  the  difference  of  two  sines. 

20.  Express  cos  5  a  cos  a  as  the  sum  of  two  cosines. 

Prove  the  following  identities  : 

21.  cos2  a  —  cos2/?  =  —  sin  (a  +  ft)  sin  (a  —  ft). 
sin  2  a  +  sin  a  3  a 


00 
22. 


-  --^ 

cos  2  a  -j-  cos  a  2 


cos  /?  —  cos  a 
24. 


1  +  tan2!  0 

25.  sin  |-  a;  +  2  sin2  ^a?  cos  J  #  cot  #  =  sin  x  cos  -|-a/*. 

26.  2  tani»csca;  =  l4-tan2^aj. 

27.  cos  (a  -f-  ft)  cos  /?  —  cos  (  «+  y)  cos  y  = 

sin  (a  -f  y)  sin  y  —  sin  (a  -f-  /?)  sin  ft. 

28.  cos2a  =  8sin4-i-a-8sin2£a  +  l. 

29.  cos  2  a  -  cos  a  =  2  (4  sin4!  a  -  3  sin2!  a). 

30.  (1  —  cos  2  a)  cot2!  a  =  cos  2  a  +  4  cos  «  +  3. 

31.  2sin2|-0cot0  —  csc20  +  csc0  =  cot2  0  +  sin0. 

32.  Given  tan  a;  =  a  tan  1  a?  ;  find  tan  ^  a;  in  terms  of  a. 

Solve  the  following  equations  for  all  positive  values  of  a 
less  than  180°: 

33.  sin2a-4cos2ia-sina  +  3  =  0. 

SOLUTION.     2  sin  a  cos  a  -  4  /1  +  cos<*\  -  sin  a  +  3  =  0, 

or  2  sin  a  cos  a  —  2  cos  a  —  sin  «  +  1  =  0, 

or  (2  cos  a  —  1)  (sin  a  —  1)  =0, 

cos  a  =  ^  or  sin  a  =  1. 
and  a  =  60°  or  90°. 


FUNCTIONS  OF  THE   SUM  OF   TWO  ANGLES     73 

34.  sin  2  a  —  cos  «  =  0. 

35.  sin  2  «  +  sin  «  =  0. 
36. 

37. 

38.  cos  3  a  +  2  cos  2  «  —  8  cos2a  +  2  sin2  a  4-  5  cos  a  =  0. 

39.  cos  2  a  H-  2  sin2!  a  —  1  =  0. 
40  sin2a  +  cos2a  +  sin  a  =  l. 

41.  sin  5  a  —  sin3cc  +  sin  a  =  0. 

SOLUTION.        2  sin  3  a  cos  2  a  —  sin  3  a  =  0.     Art.  73. 
or  sin3a(2cos2a-l)  =  0. 

sin  3  a  =  0,  or  cos  2a  =  i. 
3  a  =  0°,  180°,  360°,  and  2  a  =  60°,  300°. 
a  =  0°,  60°,  120°,  30,  150°. 

42.  cos  3  a  +  sin  2  a  —  cos  a  =  0. 

43.  sin  3  a  -+-  sin  2  «  -|-  sin  a  =  0. 

44.  cos  3  a  —  sin  2  a  +  cos  a  =  0. 

45.  sin  3  a  +  cos  3  a  —  sin  a  —  cos  a  '=  0. 

46.  sin  a;  +  sin  2  #  +  sin  3  x  =  1  +  cos  x  -f-  cos  2  #. 

47.  ^-^=20032*. 
1  +  tan  x 

Prove  the  following  identities  : 

48.  tan  (30°  +  a)  tan  (30°  -  <*)  =  2  cos  2  a  ~  1. 

'     2  cos  2  a  +  1 

49.  tan  a  -J-  cot  a  =  2  esc  2  a. 

50.  sin  80°  =  sin  40°  +  sin  20°. 

51.  1  +  sintt-cosft^^ 
1  +  sin  a  +  cos  a  2 

52.  cos2  a  -  sin  (30°  +  a)  sin  (30°  -  a)  =  f  . 


cos  «  -h  cos 


CHAPTER   VII 
INVERSE   FUNCTIONS 

76.  Inverse  trigonometric  functions  are  closely  related  to 
the  trigonometric  functions  previously  studied. 

After  introducing  the  fundamental  idea  of  an  inverse 
function,  it  will  be  shown  that  they  lead  to  new  relations, 
closely  allied  to  the  relations  already  developed. 

77.  Fundamental  idea  of  an  inverse  function.      From  the 
equation  sin  a  =  u,  it  is  evident  that  a  is  an  angle  whose 

sine  is  it.      The  statement  a  is  an 
angle  whose  sine  is  u  is  abbreviated 
into 
7  a,  =  sin"1  u. 


The  equation  u  =  sin  a  expresses 
u  in  terms  of  a. 

The  equation  a  =  sin"1  u  expresses  a  in  terms  of  u. 

We  thus  have  two  methods  of  expressing  the  relation 
between  an  angle  and  its  sine. 

The  symbol  sin"1?*  is  the  inverse  sine  of  u.  It  may  be 
read  the  inverse  sine  of  u,  the  anti  sine  of  u,  arc  sine  u,  or  an 
angle  whose  sine  is  u.  The  last  form  should  be  used  until 
the  conception  of  an  inverse  function  is  perfectly  clear. 

Corresponding  to  each  direct  function  there  is  an  inverse 
function.  Thus, 

a  =  sin"1  u  corresponds  to  sin  a  =  u, 
a  =  cos""1  u  corresponds  to  cos  a  =  u, 
a  —  tan"1  u  corresponds  to  tan  a  =  u,  etc. 

CAUTION.     Since  sin"1  u  is  an  inverse  function,  the  —  \ 

74 


INVERSE  FUNCTIONS  75 

cannot  be  an  exponent.  It  is.  simply  part  of  a  symbol  for  an 
inverse  function.  When  a  function  is  affected  by  —  1  as  an 
exponent,  it  must  be  written  with  parentheses,  thus  (sin  a;) 


-i 


78.  Multiple  values  of  an  inverse  function.  It  has  been 
shown  that  the  trigonometric  functions  are  single-valued 
functions.  Thus,  if  a  is  given,  there  is  only  one  value  for 
sin  a.  See  Arts.  25  and  41. 

On  the  contrary,  the  inverse  functions  are  multiple-valued 
functions.  Thus,  if  u  is  given  there  are  an  infinite  number 
of  values  for  sin"1^.  See  Arts.  26  and  41.  A  few  ex- 
amples will  illustrate  this  property  of  inverse  functions, 
and  show  that  all  the  values  of  any  given  inverse  function 
may  be  combined  in  a  single  expression. 

To  find  all  the  values  of  tan"1 

V3 

tan-1-lr=  30° 

V3 

=     180° +  30° 
=     360° +  30° 


-  _  180°  -f  30°,  or  - 150° 
=  _  360°  -|-  30°,  or  —  330° 
=  _5400^30°,  or  -510° 

All  these  angles  may  be  expressed   by  n  180°  +  30°,  n 
being  any  positive  or  negative  integer.     Hence 
_1   i 

VB= 

In  general,  if  a  is  an  angle  whose  tangent  is  u,  it  may  be 
shown  that  tan-i  u  =  n>n  +  Q.. 


76  TRIGONOMETRY 

To  find  all  the  values  o/cos"1^. 

cos-1^  60° 

=  -60° 

=  360°  +  60° 

=  360°  -  60° 

=  720°  +  60° 

=  720°  -  60° 


.  =  _  360°  +  60° 
=  -360°  -60° 

=  _  720°  +  60° 
=  _  720°  -  60° 


All   these   angles   may  be   expressed  by  n  360°  ±  60°,  n 
being  any  positive  or  negative  integer.     Hence 


o 

In  general,  if  a  is  an  angle  whose  cosine  is  u,  it  may  be 
shown  that  cos-i  u  =  2  nir  ±  a.. 

To  find  all  the  values  o/sin-1!. 

sin-1i=  30° 

=  180°  -  30° 

=  360°  +  30° 

=  540°  -  30° 

=  720°  +  30° 

=  900°  -  30° 


-  180°  -  30° 

-  360°  +  30° 

-  540°  -  30° 
:  -  720°  +  30° 


INVERSE  FUNCTIONS  77 

All  these  angles  may  be  expressed  by  n  180°  +  (  -  1)"30°, 
n  being  any  positive  or  negative  integer.     Hence 


In  general,  if  a  is  an  angle  whose  sine  is  w,  it  may  be 
own  that  sin-'  U  =  /nr  +(-!)"  a. 

In  a  similar  manner  it  may  be  shown  that 


sec"1  u  =  2  nir  ±  a, 
csc~lu  =  nir  +  (—  l)n  a. 

79.  Principal  values.     The  smallest  numerical  value  of  an 
inverse  function  is  called  its  principal  value,  preference  be- 
ing given  to  positive  angles  in  case  of  ambiguity. 

The  principal  values  of  the  inverse  sine  and  the  inverse 

cosecant  lie  between  —  —  and  -  ;  of  the  inverse  cosine  and 

2          2 

the  inverse  secant,  between  0  and  ?r  ;  of  the  inverse  tangent 
and  the  inverse  cotangent,  between  —  -  and  -  • 

The  principal  values  of  an  inverse  function  are  sometimes 
distinguished  from  the  general  values  by  the  use  of  a  capital 
letter. 

Thus  Sin-1-^-, 

while  sin-1  1  =  mr  +  (  -  l)w-  • 

6 

80.  To  interpret  sin  sin~lu  and  sin~l  sin  a. 

The  expression  sin  sin"1!*  is  read:  the  sine  of  the  angle 
whose  sine  is  u.  This  sine  is  evidently  u,  hence 

sin  sin'1  u  =  u. 


78  TRIGONOMETRY 

The  expression  sin"1  sin  a  is  read :  the  angle  whose  sine 
is  the  sine  of  a.  This  angle  is  evidently  a,  hence 

sin"1  sin  a  ==  a, 
or  more  generally, 

sin"1  sin  a  =  nir  -\-  (—  l)n«. 

Similar  relations  exist  between  any  direct  function  and 
the  corresponding  inverse  function. 

Thus  cos  cos"1  u  =  u ; 

cos"1  cos  a  =  a,  or  cos"1  cos  a  =  2  n-n  ±  a ; 

tan  tan"1  u  =  u  ; 

tan"1  tan  a  =  a,  or  tan"1  tan  «  =  nir  +  a,  etc. 

81.  Application  of  the  fundamental  relations  to  angles  ex- 
pressed as   inverse  functions.     The    fundamental    relations, 
being  true  for  all  angles,  must  necessarily  be  true  when 
the  angles  are  expressed  as  inverse  functions. 

Thus,  letting  a  =  tan"1  u  in  the  identity 

sin2  a  +  cos2  a  =  1, 

we  have  (sin  tan"1  u)2  -f-  (cos  tan"1  u)2  =  1. 

Similarly         (sin  esc"1  u)2  -f-  (cos  esc"1  u)2  =  1 ; 

(tan  sec"1  u)2  +  1  =  (sec  sec"1  u)2  j 

sin  cos"1  u  = •  • 

esc  cos"1  u 

By  expressing  the  angle  of  the  fundamental  relations  as 
an  inverse  function,  we  may  develop  relations  between  the 
inverse  functions. 

82.  Given  an  angle,  expressed  as  an  inverse  function  of  u, 
to  find  the  value  of  any  function  of  the  angle  in  terms  of  u. 

By  the  application  of  one  or  more  of  the  fundamental 
relations,  it  is  always  possible  to  solve  the  stated  problem. 


INVERSE  FUNCTIONS 


79 


Several  illustrations  are  given  below.  The  method  em- 
ployed can  be  readily  applied  to  the  other  functions. 

1.    To  find  the  value  of  tan  cos"1  u  in  terms  of  u. 

If  tan  cos"1  u  is  expressed  in  terms  of  the  cosine  of 
cos"1  u,  the  problem  is  solved,  since  cos  cos"1  u  =  u. 

This  may  be  done  as  follows  : 


tan  cos"1  u 


sin  cos"1  u  _  ±  Vl  —  (cos  cos"1  u)2  __  ±  Vl  —  u2 

COS  COS"1  U  U  U 


This  result  may  be  obtained  geometrically.  Since  u  is 
given,  it  is  evident  that  cos"1^  represents,  among  others, 
two  positive  angles,  a±  and  «o,  each  less  than  360°. 

Let  us  assume  u  positive  and  let  us  construct  these  angles 
defined  by  cos"1  u. 

Then  from  the  figure  and  the  definition  of  the  tangent, 


and 


Since  the  tangent  of  any  angle  Co- 
terminal   with   «!  is  -t- —      — ,  and  the  tangent  of  any 


angle  coterminal  with  0%  is    — — — ,  and  since  cos"1^ 

u 

represents  all  angles  coterminal  with  either  ai  or  #2,  we  have 


tan  cos"1  u  = 


±  Vl  -  u* 


2.    To  find  the  value  of  sec  cot  1  u  in  terms  of  u. 

To  solve  this  problem  it  is  only  necessary  to  express 
sec  cot"1  u  in  terms  of  cot  cot"1  u. 


80 


TRIGONOMETRY 


Thus 


sec  coir1  u  =  ±  Vl  4- (tan  cot"1  u)2  =  -v/1  4- 


(cot  cot"1  w)2 


This  result  may  be  obtained  geometrically.  Construct 
the  angles  given  by  cot"1  u.  Let  us  assume  in  this  problem 
that  16  is  negative  and  hence  that  —  u  is  positive.  If  the 

cotangent  of  an 
^angle  is  nega- 
tive the  angle 
must  terminate 
in  either  the 
second  or  fourth 
quadrant.  Since 
OA  and  OB  are 

the  terminal  lines  of  otj  and  «2  respectively,  and  since  the 
terminal  line  is  always  positive,  we  have 
OA  =  OB  = 


u  —u 

or  by  considerations  similar  to  those  in  the  previous  ex- 
ample, we  have 

+  Vl  4-  u2      ±  Vl  4-  w2 


sec  cot"1 16=- 

±16  U 

3.  To  find  the  value  of  sin  cos"1  u  in  terms  ofu. 

We  have  sin  cos"1  u  =  ±  Vl  —(cos  cos"1 16)2  =  ±  Vl  —  i*2. 

4.  To  find  the  value  of  cot  vers"1 16  in  terms  of  u. 
We  have 


1-u 


±  Vl  —  (1  —  vers  vers"1  w)2 
1-u 

±^/2u  —  u* 


INVERSE  FUNCTIONS  81 

83.   Some  inverse  functions  expressed  in  terms  of  other  in- 
verse functions. 

1.  To  express  cos"1  u  in  terms  of  an  inverse  tangent. 

From   tan   cos"1  u  =  — ,     (Art.   82,  prob.   1)  by 

u 

taking  the  inverse  tangent  of  each  member  (Art.  80),  there 

results  ±  yi  _  ^2 

cos"1  u  =  tan"1 —  • 

u 

2.  To  express  the  cot"1  u  in  terms  of  an  inverse  secant. 


From  sec  cot"1  u  =  ±        +  M>     (Art.  82,  prob.  2)  there 


u 


results 


3.  Similarly  from  sin  cos"1  u  =  ±  Vl  —  w2  there  results 

cos"1  u  =  sin"1  (  ±  Vl  —  w2). 

4.  Also  from  cot  vers"1  M  = ~  there  results 

±  V2  u  -  u* 
1-u 


vers"1  u  =  cot"1 


±  V2  u  -  u2 

By  the  method  exemplified  in  Arts.  82  and  83  it  is  possi- 
ble to  express  any  inverse  function  in  terms  of  any  other 
inverse  function. 

In  applying  the  above  formulas  care  must  be  exercised 
in  selecting  the  angles,  since  each  inverse  function  repre- 
sents an  infinite  number  of  angles  and  one  member  of 
the  equation  may  represent  angles  not  represented  by  the 
other.  For  example,  in  problem  1,  if  u  be  positive  the 
cos"1  u  represents  angles  terminating  in  the  first  and  fourth 

quadrants  ;  but  tan"1 -5^- — -^-  represents  angles  terminat- 
ing in  the  second  and  third  quadrants  as  well  as  angles 
terminating  in  the  first  and  fourth  quadrants. 


82  TRIGONOMETRY 

84.  Some  relations  between  inverse  functions  derived  from 
the  formulas  for  double  angles,  half  angles,  and  the  addition 
formulas. 

The  general  method  applicable  to  this  class  of  problems 
will  be  illustrated  by  a  few  examples. 

1.    To  express  cos  (2  sec"1  u)  in  terms  of  u. 

cos  (2  sec"1  u)  =  2  (cos  sec"1  u)z  -  1     Art.  71,  Eq.  4. 

1=1-1. 


(sec  sec"1  u)2  u2 

From  this  relation  it  follows  that, 

(2        \ 
1 ). 
u2        ) 

2.    To  express  tan  (1  cos"1  u)  in  terms  of  u. 

lY^^L 


COS 

Art.  72,  Eq.  3. 
From  this  relation  it  follows  that, 

l ,         *. ,,  _i_    /I  —  if 

2 


3.    To  express  sin  (sin"1  u  +  cos"1  v)  in  terms  of  u  and  v. 
sin  (sin"1  u  +  cos"1  v)  =  sin  sin"1  u  •  cos  cos"1  v 
+  cos  sin"1  u  •  sin  cos"1  v 

Art.  62,  Eq.  1. 


From  this  relation  it  follows  that, 


sin"1  u  +  cos"1  v  =  sin"1  (uv  ±  Vl  —  w2  Vl  —  v2). 

85.  EXAMPLES 

Find  the  value  of  each  of  the  following : 

1.  sin-^VS.  3.   tan"1!. 

2.  cos-^-iVS).  4.   tan  cot"1 4. 

5.   sin  cot"1 4. 


INVERSE  FUNCTIONS  83 

Express  the  following  in  terms  of  u  and  v: 

6.  cos  cot"1  w.  17.   cos  (2  cos'1  w). 

7.  sec  cot"1  u.  18.   cos  (2  skr1  u). 

8.  esc  cot"1  u.  19.   cos  (2  tan"1  u). 

9.  cos  sin"1  u.  20.   sin  (sin"1  u  -f  sin"1  v). 

10.  cos  tan"1  u.  21.  cos  (sin"1  w  4-  sin"1  v). 

11.  cossec^w.  22.  tan  (tan"1  u  +  cot"1  v). 

12.  cos  esc"1  u.  23.  tan  (sec"1  u  +  sec"1  v). 

13.  sin  (2  cos"1  w).  24.  cos  (see"1  w  +  osc"1  v). 

14.  tan  (2  tan-1  w).  25.  sin  (£  cos-1  w)« 

15.  tan  (2  sec-1  w).  26.  cos  (J  cos'1  u). 

16.  tan  (2  cos-1  w).  27.  sin  (i  sec'1  w). 

28.    sin  (|  tan-1  w). 
Find  x  in  terms  of  a. 

29.  tan"1  a;  =  cot"1  a. 

30.  sin  -1  x  =  tan"1  a. 

31.  cos"1  x  =  2  sin""1  a. 

32.  cos"1  #  =  sin"1  a  +  tan"1  a. 

33.  sin"1  x  =  -J-  sec"1  a. 
Find  a;  in  terms  of  a  and  &. 

34.  tan"1  x  =  sin"1  a  +  sin"1 6. 

35.  cos"1  x  =  sec"1  a  —  sec"1  b. 

36.  sin"1  x  =  2  cos"1  a  -f-  £  cos"1 6. 


CHAPTER   VIII 
OBLIQUE    TRIANGLE 

86.  In  the  present  chapter  we  develop  the  formulas  by 
means  of  which  a  triangle  may  be  completely  solved  when 
any  three  independent  parts  are  given. 

87.  Law  of  sines.     In  a  plane  triangle  any  two  sides  are  to 
each  other  as  the  sines  of  the  opposite  angles. 


c  c 

Let  a,  b,  c  be  the  sides  of  a  triangle  and  a,  ft,  y  the  angles 
opposite  these  sides,  respectively. 

From  the  vertex  of  y  draw  h  perpendicular  to  the  side  c, 
or  c  produced. 


Then,  for  each  figure,   sin  a  =  - , 

b 


and 


siny3=-< 
a 


Art.  21 
Art.  29 


Dividing  the  first  equation  by  the  second,  we  have 

a  _  sin  a 

b      sin/? 

In  a  similar  manner,  dropping  a  perpendicular  from  the 
vertex  of  a,  it  is  seen  that 


c      sny 
The  last  two  equations  may  be  written 

a  b  c 

sin  a     sin  p      sin  -y 
84 


OBLIQUE   TRIANGLE  85 

88.   Law  of  tangents.     The  tangent  of  half  the  difference  of 
two  angles  is  to  the  tangent  of  half  their  sum,  as  the  difference 
of  the  corresponding  opposite  sides  is  to  their  sum. 
From  the  law  of  sines  we  have 
a  _  sin  a 
b      sin/3 

By  division  and  composition,  this  becomes 
a  —  b  _  sin  a  —  sin  ft 
a  +  b      sin  a  +  sin/?' 

which  reduces  to 

a-  6     2cos   -(«  +      sina—  ff) 


a  +  6     2  sin  1  (a  +  J3)  cos  |  (a  -  ft 
=  cot  i  (a  +  J3)  tan  *-  («  -  J3) 


tana  +          a  + 

Singly  tanHP-V)      ^_g,  (2) 

tani(p+Y>      6+* 
and  tanj^^Lo)  =  c--a>  (3) 

tan  i  (-y  +  a)      c  +  a 

89.  Cyclic  interchange  of  letters.     Each  formula  pertaining 
to  the  oblique  triangle  gives  rise  to  two  other  formulas  of 
the  same  type  by  a  cyclic  interchange  of 

letters.     A   cyclic   interchange   of  letters 

may  be  accomplished    by   arranging    the 

letters  around  the  circumference  of  a  circle 

as  in  the  figure,  and  then  replacing  each 

letter  by  the  next  in  order  as  indicated  by 

the   arrows.     Thus   by   this   cyclic   interchange  of  letters 

formula  (1)  of  Art.  88  gives  rise  to  formula  (2)  ;  likewise 

formula  (2)  gives  rise  to  formula  (3). 

90.  Law  of  cosines.     The  square  of  any  side  of  a  plane  tri- 
angle is  equal  to  the  sum  of  the  squares  of  the  other  sides 
minus  twice  the  product  of  those  sides  into  the  cosine  of  the 
included  angle. 


86 


TRIGONOMETRY 


c    D 


B 


DA  B 

For  each  figure 

AB  =  AD  +  DB,     or     DB  =  AB-  AD.      Art.  2 
But          AB  =  c,  ^4Z>  =  6  cos  a,     hence  DB  =  c  —  b  cos  a. 
Also  D(7  =  b  sin  a. 

From  the  right  triangle  CDB  we  have 

BC2  =  DC2  -\-  DB2 

Substituting  values,  this  equation  becomes 
a2  =  (6  sin  a)2  +  (c  —  6  cos  a)2 

=  62sin2a  +  c2  —  2  fcccos  a  +  62cos2a 
=  62(sin2  a  +  cos2  a)  +  c2  —  2  6c  cos  a. 
Hence  a2  =  62  +  c2  -  2  be  cos  a. 

Similarly        62  =  c2  +  a2 
and  c  ==  fl  ~f~  o  • 

91.    To  find  the  sine  of  half  an  angle  of  a  plane  triangle  in 
terms  of  the  sides  of  the  triangle. 

From  equation  (1),  Art.  72, 


—  cos  « 


whence 


From  the  cosine  law 


.  cos  a.  = 


52  +  c2  _ 
2  be 


(1) 


(2) 


OBLIQUE   TRIANGLE  87 

• 

From  equations  (1)  and  (2) 


2  be 


2  be 
(a  —  6 


2  be 
Let  a  +  b  -f  c  =  2  s.  (4) 

Subtracting  2  a,  26,  and  2  c  from  each  member  of  (4)  we 
have  respectively 

-a  +  6  +  c  =  2(S-a) 

a  -f-  6  —  c  =  2  (s  —  c). 
Then  equation  (3)  becomes 


a 

Similarly  sin  •£  =  -j- 


ca 


-82. 

^/in  ter 


and  6in?  =  +  J(* -")(*-*). 

2          *  a0 

In  these  formulas  the  positive  sign  is  given  to  the  radical, 
since  it  is  known  that  half  an  angle  of  any  plane  triangle  is 
less  than  90°.  The  same  applies  to  the  corresponding  for- 
mulas for  the  cosine  and  the  tangent  of  half  an  angle. 

To  find  the  cosine  of  half  an  angle  of  a  plane  triangle 
'n  terms  of  the  sides  of  the  triangle. 


88 


TRIGONOMETRY 


From  equation  (2),  Art.  72,  we  have 

2cos2-  =  l  +  cosa. 
2 


Then 


1    .  fr2  +  c2  -  a2 
l+___ 

(b  +  c)2  -  a2 
2  be 


26c 


and 
Hence 

Similarly 
and 


93.    To  find  the  tangent  of  half  an  angle  of  a  plane  tri- 
s    angle  in  terms  of  the  sides  of  the  triangle. 


Since 

Similarly 
and 

sin? 
tan"-        2- 

(1) 
(2) 
(3} 

HUI 

A         ,        a 

tan  a  —  +  •%  fe 

_  £)($  _  c) 

2  ""      \ 

5(5  -  a) 

tan  P  —  4-  A  /(5 

-  c)(s  —  a) 

2            * 

s(s  -  b) 

,nY-        ^fe 

-a)(s-b)t 

OBLIQUE   TRIANGLE 
Formula  (1)  may  be  written 


89 


Letting 

Similarly 
and 

tan"        1    ,/(«-«)(*-&)(*  -c>- 

(4) 
(5) 
(6) 

(7) 

Ln2     s-a\ 

r_:,_.J(«-«)(*-W*-c) 

*                  * 
tana-     r 

2     s  -a 

tanp-     r 

2~«-6' 
tan''         r 

2     s-o 

94.    To  find  the  area  of  a  plane  triangle  in  terms  of  two 
sides  and  the  included  angle. 


Let  A  be  the  area  and  h  the  altitude.    Then  in  each  figure 

A  =  ^  ch, 

and  h  =  b  sin  a. 

Therefore  A  =  \  be  sin  a. 

Similarly  A  =  J  ca  sin  0, 

and  A  =  \ab  sin -y. 

95.    To  find  the  area  of  a  plane  triangle  in  terms  of  a  side 
and  two  adjacent  angles. 

From  Art.  94,  A  =  %  be  sin  a. 


90  TRIGONOMETRY 

From  the  sine  law      ,  _  csin/? 

siny 
.  _  c2  sin  a  sin  J3 

2  siny 
Then  since       a  +  ft  +  y  =  180°, 

M  _c2  sin  a  sin  p 
~2sin(a  +  p)' 

96.    To  find  the  area  of  a  plane  triangle  in  terms  of  the 
three  sides. 

From  Art.  94. 

A  =  ±bc  sin  a. 

Since  sin  a  =  2  sin  -  cos  - . 

2       2' 

A  — be  sin  -  cos  -  • 

2        2 

Substituting  the  values  of  sin  ^  and  cos  ^  as  found  in 
Arts.  91  and  92,  we  have,  after  reduction, 


-a)(S-b)(s-c). 

97.  Formulas  for  solving  an  oblique  triangle.  The  formu- 
las developed  in  the  present  chapter  are  sufficient  to  solve  a 
plane  triangle  when  three  independent  parts  are  given. 

The  law  of  sines 

a  b  c 


sin  a     sin  ft      sin  y 

is  used  when  two  of  the  given  parts  are  an  angle  and  the 
opposite  side. 

The  law  of  tangents 

tan  i  (a-  ft)  =  ^|  tan  1  (a  +  0) 

a  +  6 

is  used  when  two  sides  and  the  included  angle  are  given. 


OBLIQUE   TRIANGLE  91 

If  a,  6,  and  y  are  the  given  parts,  \  (a  +  ft)  is  obtained  from 
the  relation  a  +  /?  +  y  =  180°.  The  formula  then  gives  the 
value  of  -J  (a  —  /3),  which,  united  with  the  value  of  %  (a 
gives  a  and  /?. 

TTie  ta;s  of  half  angles 


s  (s  —  a)         s  —  a 

are  used  when  the  three  sides  are  given.  The  last  formula 
is  the  most  accurate,  since  the  tangent  varies  more  rapidly 
than  either  the  sine  or  the  cosine.  The  formula  involving 
r  is  advantageous  when  all  the  angles  are  to  be  computed. 

TJie  law  of  cosines 


may  be  used  to  determine  the  third  side  when  two  sides 
and  an  angle  are  given.  It  may  also  be  used  to  determine 
an  angle  when  the  three  sides  are  given. 

This  formula  is  used  with  natural  functions,  not  being 
adapted  to  logarithmic  computation. 

98.  Check  formulas.  Any  formula  which  was  not  used 
in  the  solution  of  the  triangle  may  be  used  as  a  check 
formula. 

The  relation  a  +  (3  +  y  =  180°  cannot  be  used  as  a  check 
when  a  problem  has  been  solved  by  the  law  of  tangents, 
since  the  law  of  tangents  involves  this  relation. 

When  two  equations  from  the  sine  law  have  been  used  to 
find  two  elements  of  a  triangle,  the  third  equation  from  the 
sine  law  cannot  be  used  as  a  check,  since  the  first  two 
equations  involve  the  third. 


92 


TRIGONOMETRY 


99.   Illustrative  problems. 

1.   Given  c  =  127.32 

a  =  71°  58'  22" 
J3  =52°  19'  40" 

SOLUTION.     Construction  and  estimates. 

A 
A\ 

7      V 


to  find  a 
b 
y. 


a  =  140 

6  =  120 

7  =  60° 


••A 


Outline 


c  sin  a                       tan  *  fa     ^      a 

C%ecfc 
_  5 

ian  2  ^«      p; 
sm  7                                                      a 

+  6 

c 
a 
ft 

a-6 

a  +  6 

y 

log  a  —  b  ) 
colog   (a  +  6) 

log  sin  a 

log  tan  £  (a  +  /3) 

logc 
colog  sin  7 

log  tan  £  (a  —  /3) 

log  a 
a 

!>+« 

, 

~ 


sin  7 


r,  more  compactly, 


log  sin  /3 
logc 
colog  sin  7 
log& 
6 

log  sin  a 

.  w>  j  colog  sin  7 
^    [      log  sin  /3 
log  a 
log& 
a 
6 


OBLIQUE  TRIANGLE 


93 


Filling  in  the  above  outline,  the  completed  work  appears  as  follows 


c  sin  a 


sin  7 

c 
a 

V 

127.32                                                    c,     , 
71°  58'  22" 
52°  19'  40"                      tan*Ca    /3)      a  ~  &  tan  *fa  +  /3) 

124°  18'  2"                                                a  +  6 

55°  41'  58"                                          a-6 

24.57 
268.55 
62°  9'  1" 

log  sin  a 
logc 
colog  sin  7 
log  a 
a 

9.97814-10                                        a  +  b 
2.10490                                         ?(«  +  £) 

0.08297                                       log  (a  -6) 

1.39041 
7.57098 
0.27708 

2.16601                                     colog  (a  +  6) 
146.56                               log  tan  £  (a  +  /3) 

3  sin/*                                logtanKa-0) 

9.23847-10 
9°  49'  28" 
62°  9'  1" 
71°  58'  29" 
52°  19'  33" 

Shl7                                              Ha  -I-/9) 

log  sin  ft 
logc 
colog  sin  7 
log  6 
6 

Or,  in  the 

9.89846-10                                            '  r'a 
2.10490                                                         ,3 
0.08297 

2.08633 
121.99 

more  compact  form, 

(log  sin  a     9.97814-10 
f           logc     2.10490 
"S)j  colog  sin  7     0.08297 
-2  1      log  sin  ft     9.89846 

log  a     2.16601 
log  b     2.08633 
a     146.56 
b     121.99 

2.    Given 


a  =  1674.3 
c  =  1021.7 
£  =  28°  44'  39" 


to  find  a 

7 
b. 


Estimates 
a  =  120° 
7  =  30° 
6  =  900. 


94 


TRIGONOMETRY 


a  -f  c 


T  _  c  sin  /3 
sin  7 


a 
c 

/s 

a  —  c 
a  +  c 
a  +  7 


log  (a  -  c) 

colog  (a  +  c) 

log  tan  £  (a  +  7) 

log  tan  H«-7) 


1674.3 
1021.7 
28°  44'  39" 


652.6 
2696.0 
151°  15'  21" 

75°  37'  40" 


logo 

log  sin  /3 

colog  sin  7 

log  6 


3.00932 

9.68205-10 

0.27268 


2.96405 
920.66 


Check 


2.81465 

6.56928-10 

0.59135 


b  = 


sin  a 


9.97528-10 
43°  22'  12" 
75°  37'  40" 


118°  59'  52" 
32°  15'  28" 


log  a 

log  sin  0 

colog  sin  a 

log  6 


3.22384 

9.68205-10 

0.05817 


2.96406 


3.    Given 


a  =  1.4932 
b  =  2.8711 
=  1.9005 


to  find 


a 

Estimates 

ft 

a=    25° 
/3  =  120° 

y- 

7  =    35° 

r  — 


tin7           r 

2       S  -  C 

a 

1.4932 

6 

2.8711 

c 

1.9005 

2s 

6.2648 

s 

3.1324 

s  —  a 

1.6392 

s-  6 

0.2613 

*  —  c 

1.2319 

s 

3.1324 

log  (s  -  a) 

0.21463 

log  0-6) 

9.41714  -  10 

log  (s  -  c) 

0.09058 

colog  s 

9.50412  -  10 

logr2 

19.22647  -  20 

logr 

9.61324  -  10 

log  tan  J 

9.39861  -  10 

log  tan  | 

0.19610 

log  tan  J 

9.52266  -  10 

• 

2 

14°    3'  26" 

e 

2 

57°  31'   2" 

2 
2 

18°  25'  34" 

a 

28°   6'  52" 

ft 

115°   2'   4" 

7 

36°  51'    8" 

CAecfc  :  a  +  0  +  7 

180°  00'    4" 

OBLIQUE   TRIANGLE 


95 


4.  Given 


6  =  .0060041 
c  =  .0093284 
=  44°  47' 58" 


to  find 


Estimates 
j8=  30° 
a  =  105° 
a  =  .012. 


Check 


sin/3 

6  sin  7 

tin  ""  ("ft      "v}  —  a       c  tan  1  (OL  1  V^ 

c 

a  +  c 

b 
c 
7 

.0060041 
.0093284 
44°  47'  58" 

a 
c 
a  —  c 

a  +  c 
a  +  7 
K«  +  7) 

.012574 
.0093284 
.0032456 
.021902 
153°    1'48" 
76°  30'  54" 

log& 
colog  c 
log  sin  7 
log  sin  /3 

a 
a 

7.77845  -  10 
2.03019 
9.84796  -  10 

9.65660  -  10 
26°  58'  12" 
71°  46'  10" 
108°  13'  50" 

csina 

log(a-c) 
colog  (a  +  c 
log  tan  Ka  +  7) 
log  tan  £  («  -  7) 
U«-7) 

7 

7.51129  -  10 
1.65952 
0.62015 

9.79096  -  10 
31°  42'  51 
76°  30'  54" 
108°  13'  45" 
44°  48'    3" 

sin  7 

logc 
colog  sin  7 
log  sin  a 
log  a 
a 

7.96981  -  10 
0.15204 
9.97764-10 

8.09949-10 
0.012574 

100.  The  ambiguous  case.  When  two  sides  and  an  angle 
opposite  one  of  them  are  given,  the  triangle  may  admit  of 
no  solution,  of  one  solution,  or  of  two  solutions. 

Let  a,  b,  a  be  given.     The  formula 


sin/?; 


determines 


96 


TRIGONOMETRY 


If  the  calculated 
sin  ft  is  greater  than  1, 
there  can  be  no  solu- 
tion. 


a  <  b  sin  a 


If  the  calculated 
sin  ft  equals  1,  ft  = 
90°  and  there  is  one 
solution. 


.5 1  a 


a  =  6  sin  a 


If  the  calculated 
sin  ft  is  less  than  1, 
two  supplementary 
values  of  ft  are  deter- 
mined, giving  two  so- 
lutions unless  the 
larger  value  of  ft  plus 
a  is  equal  to  or  greater 
than  180°. 


a  >  b  sin  a 

a<b 


Given 


6  =  420 
c  =  389.73 
y  =53°  47'  20" 


to  find  ft 
a 
a. 


Estimates 

/3  =  65° 
a  =60° 
a  =  390. 


Ttoo  solutions 

/S'  =  115° 
a'  =  10° 
a' =  90. 


a 


OBLIQUE   TRIANGLE 


97 


Ch'jck  of  1st  solution 


sin/3  = 


6  sin  7 


tan  \  (a  -  7)  = 


-  - 
a  +  c 


b 
c 
y 

420 
389.73 
53°  47'  20" 

a 
c 
a  —  c 

a  +  c 
ft  +  7 
*(«  +  7) 

440.61 
389.73 
50.88 
830.34 
119°  35'  51' 
59°  47'  50" 

log  b 
colog  c 
log  sin  7 
log  sin  p 
ft 
180°  -p  =  p' 
P  +  y 

£'  +  7 
ft 
ft' 

2.62325 
7.40924  -  10 
9.90679  -  10 

9.93928  -  10 
60°  24'    9" 
119°  35'  51" 
114°  11  '29" 
173°  22'  11" 
65°  48'  31" 
6°  37'  49" 

log  (a  -  c) 
colog  (a  +  c) 
log  tan  £  («  +  7) 
log  tan  £  (a  -  7) 

i(«-7) 

H«+7) 

ft 

7 

1.70655 
7.08074-  10 
0.23502 

9.02231  -  10 
6°   0'34" 
59°  47'  50" 
65°  48'  24" 
53°  47'  16" 

Check  of  2d  solution 


tan  \  (7  -  a')  = 


c  +  a 


tan  |  (7  + 


c  sin  ft 

c 
a' 
c  —  a' 
c  +  a' 
7  +  ft' 

*  (y  +  «f) 

389.73 
55.771 
333.96 
445.50 
60°  25'  9" 
30°  12'  34" 

sin  7 
,      c  sin  a' 

sin  7 

log  sin  a 

r      logo 

II    colog  sin  7 
log  sin  a' 
log  a 
log  a' 
a 
a' 

9.96008  -  10 
2.59076 
0.09321 
9.06244  -  10 

log  (c  -  a') 
colog  (c  +  a') 
log  tan  i  (7  +  a') 
log  tan  £  (7  -  a') 
H7-ft') 

K7  +  «0 
7 
ft' 

2.52370 
7.35115  -10 
9.76509-10 

9.63994  -  10 
23°  34'  45" 
30°  12'  34" 
53°  47'  19" 
6°  37'  49" 

2.64405 
1.74641 
440.61 
55.771 

FIRST    SOLUTION 

p  =  60°  24' 9" 
ft  =  65°  48' 31" 
a  =  440.61 


SECOND   SOLUTION 

j8'  =  119°35'51" 
a*=  6°  37' 49" 
a' =  55.771 


98  TRIGONOMETRY 

101.  EXAMPLES 

Solve  the  following  triangles,  using  a  three-place  table  or 
a  slide-rule. 

1.   a  =26  2.   a  =  48 

a  =  53°  ft  —  61° 

ft  =  49°  y  =  69° 

3.    a  =  73°  4.   a  =  69° 

6  =  55  6  =  64 

5.    a  =  54°  40'  6.   a  =  163 

6  =  122  6  =  241 

c  =  110  y  =  34°  20' 

7.   a  =  42  8.    6  =  115 

6  =  28  c  =  96 

y  =  72°  a  =  110° 

9.   a  =  51  10.  a  =  8.03 

6  =  63  6  =  6.42 

c  =  70  c  =  7.15 

11.   a  =  20°  12.   y  =  43°50' 

a  =  15  a  =  .34 

6  =  25  c  =  .30 

Solve  the  following  triangles : 

^13.    6  =  63.67  14.    6' =20.007 

ft  =  100°  10'  a  =  40°  27'  30" 

«  =  40°0'10"  y  =  42°  30' 15" 

15.   a  =  238.61  . NsJ6.   a  =  8.0038 

6  =  216.77  6  =  4.6259 

c  =  98.435  c  =  4.3167 

17.    6  =  .76328  ^4.8-    6  =  85.249 

c  =  2.4359  c  =  105.63 

y  =  120°  46'  18"  a  =  50°  40'  24" 


OBLIQUE   TRIANGLE 


99 


19.  a  =  1.4562 
c  =  .45296 
/3  =  74°  19' 38" 

21.  a  =  2.1469 
6  =  3.2824 
j^=  4.0026 

23.  6  =  .06532 
c  =  .01846 
y  =  8°  0'  20" 

25.;  a  =  764.38 

«  =  143° 18'  31" 
13  =  13°  34'  26" 


20.   a  = 

6  = 
c  = 

22.   c 

a 
a 

24.   6 

c 

26.   a 

6 

c 


83.831 
56.479 
74.025 

7.2693 
.54871 
5°  41'  30" 

10.246 
18.075 
33°  30' 5" 

962.27 
637.34 
655.80 


Find  the  areas  of  the  following  triangles 


28.  a  = 

c  = 

30.  a  = 

6  = 

\32.  c  = 

a  = 

^54.  6  = 


27.  a  =  15 

6  =  20 
c  =  25 

29.  a  =  20.46 
6  =  19.72 
c  =  15.04 

31.  6  =  3.46 
c  =  4.09 
a  =  56°  10' 

33.  a  =  48 

v  =  43° 


172 
103 
141 

18.3 
22.4 
32° 

435.3 

289.6 
31°  7' 

10.34 

83° 22' 
60°  40' 


35.  The  horizontal  distance  from  a  point  on  top  of  a 
tower  to  a  distant  flagpole  is  468  ft.  The  angle  of  elevation 
of  the  top  of  the  flagpole  is  5°  8'  30.".  The  angle  of  depres- 
sion of  the  foot  of  the  pole  is  15°  36'.  Find  the  height  of 
the  flagpole. 


100  TRIGONOMETRY 

36.  A  tower  is  situated  on  a  hill  which  inclines  at  an 
angle  of  23°  19'  10"  to  the  horizontal.     The  angle  of  eleva- 
tion of  the  top  of  the  tower,  from  a  point  on  the  hillside, 
was  measured  and  found  to  be  43°  39'  50".     At  a  point  75.5 
ft.  farther  down,  the  angle  of  elevation  was  found  to  be 
39°  23'  20".     How  high  is  the  tower  ? 

37.  From  a  point  5  miles  from  one  end  of  a  lake  and 
3  miles  from  the  other  end,  the  lake  subtends  an  angle  of 
47°  34'  30".     Find  the  length  of  the  lake. 

38.  Find  the  altitudes  of  a  triangle  whose  sides  are  125.4, 
230.6,  and  179.8. 

39.  A  flagpole  stands  on  the  summit  of  a  hill.     The  hill 
inclines  32°  18'  20"  to  the  horizontal.     At  a  point  25.5  ft. 
from  the  base  of  the  flagpole,  measured  along  the  incline, 
the  angle  subtended  by  the  flagpole  was  found  to  be  41°  24'. 
Find  the  height  of  the  flagpole. 


Two  observers,  A  and  B, 
stationed  4000  ft.  apart,  at  the  same 
instant  observe  the  angles  BAG  and 
CBA  to  an  automobile  traveling  on 
a  straight  road.  Three  minutes 
later  they  measure  the  angles  DAB  and  ABD  to  the  second 
position  of  the  automobile. 


If 

Z  CBA  =  32°  8', 
2.  DAB  =  40°  12', 


what  is  the  rate  of  the  automobile  ? 

41.  A  man  wishes  to  measure  the  length  of  a  lake  from 
his  position  on  a  hill  top  185  ft.  above  the  level  of  the  lake. 
He  finds  the  angles  of  depression  of  the  ends  of  the  lake  to  be 
6°  18'  and  2°  30'.  The  angle  subtended  by  the  lake,  formed 


OBLIQUE   TRIANGLE 


101 


by  the  two  lines  of  sight,  is  66°  27'. 
the  lake. 


Find  the  length  of 


(42)  The  angles  of  elevation  of  a  cloud,  directly  above  a 
straight  road,  from  two  points  of  the  road  on  opposite  sides 
of  the  cloud,  are  78°  15'  20"  and  59°  47'  40".  Find  the  height 
of  the  cloud,  the  distance  between  the  two  points  of  observa- 
tion being  5000  ft. 

43.  Two  observers  at  A  and 
B.  whose  longitudes  are  the 
same,  simultaneously  observe 
the  moon  and  find  the  angles 
ZAM&vdi  Z'BMto  be  35°  2'  20" 
and  51°  17'  10"  respectively. 
The  latitude  of  A  (Greenwich) 
is  N,  51°  17'  15"  and  the  lati- 
tude of  B  (Cape  of  Good  Hope)  is  S.  33°  45'  16".  The  moon 
is  in  the  plane  determined  by  A,  E}  and  B.  Find  EM,  the 
distance  from  the  center  of  the  earth  to  the  moon,  the 
radius  of  the  earth  being  3960  miles. 

44.  Show  that  twice  the  radius,  2  Jf2,  of  a 
circle  circumscribing  a  triangle  is.  given  by 
the  equations 

2  £  _     a     _     b     _     c 

sin  a      sin  ft      sin  y 
SUGGESTION.     ZBAC=£DOC. 
45.    Find  the   radius   of  a  circle  in- 
scribed in  a  triangle  whose   sides   are 
given. 

SOLUTION.     Representing  the  area  of  the 
triangle  by  A,  we  have 

A  —  \ ;  ar  +  £  br  +  J  cr  =  rs. 

But  A  =  Vs(s  -  a) («-&)(*-  c).     Art.  96. 

Therefore        r  =J(*  -  «)(*  ~  *»)(*  -  <0.     See  Art.  93. 

\  s 


MISCELLANEOUS   EXERCISES 

102.  1.  An  angle  of  4  radians,  having  its  vertex  at  the 
center  of  a  circle,  intercepts  an  arc  of  7  inches;  find  the 
radius  of  the  circle. 

2.  Express  the  following  in  degrees :  \  radians,  -^  radi- 

o  o 

STT       ,, 

ans,  -j-  radians. 

3.  Express  the  following  in  radian  measure:  40°,   55°, 
38°,  52°  16'. 

4.  Eeduce  f  radians  to  degrees. 

5.  Find  the  number  of  degrees  in  a  central  angle  which 
intercepts  an  arc  of  5  feet  in  a  circle  whose  radius  is  8  feet. 

6.  An  arc  of  12  inches  subtends  a  central  angle  of  50° ; 
find  the  radius  of  the  circle. 

7.  The  number  of  minutes  in  an  angle  is  1\  times  the 
number  of  degrees  in  its  supplement ;  find  the  number  of 
radians  in  the  angle. 

8.  An  angle  exceeds  another  by  ^  radians,  and  their  sum 

is  160° ;  express  each  angle  in  radian  measure. 

9.  The  angles  of  a  triangle  are  to  each  other  as  2 : 3 : 4 ; 
express  each  angle  in  radian  measure. 

10.  The  angles  of  a  triangle  are  in  arithmetical  progres- 
sion, and  the  mean  angle  is  twice  the  smallest ;  express  each 
angle  in  radian  measure. 

102 


MISCELLANEOUS  EXERCISES  103 

11.  The  circumference  of  a  circle  is  divided  into  7  parts 
in  arithmetical  progression,  the  greatest  part  being  10  times 
the  least ;  express  in  radians  the  angle  which  each  arc  sub- 
tends at  the  center. 

12.  An  arc  of  40°  on  a  circle  whose  radius  is  6  inches  is 
equal  in  length  to  an  arc  of  25°  on  another  circle ;  what  is 
the  radius  of  the  latter  circle  ? 


Prove  each  of  the  following : 

13.  sin  «  cos  a  =  sin3  a  cos  a  -f  cos3  a  sin  a. 

14.  cot  a  esc  a  = . 


15. 


sec  a  —  cos  a 
1  +  tan2  a  _  sin2  a 
1  +  cot2  a   cos2  a 


1/S         •       a    ,          SID.  ft  COS  ft  0 

16.  sm£  +  -        ^      =  —       ___cos& 

cot  ft  —  1      1  —  tan  ft 

17.  sin6  a  -\-  cos6  a  =  1  —  3  sin2 «  cos2  a. 

18.  cot  a  —  tan  cc  ==  esc  a  sec  a  (1  —  2  sin2  a). 

19     1  ~  tanfi  _  Got  ft -I 
1  +  tan  £  ~~  cot/2-f  l' 

20.  sec4  a  —  tan4  a  =  2  sec2  a  —  1. 

21.  tan  15°  =  2 -V3. 

22.  sin  4  a;  =  4  (cos3  cc  sin  a?  —  sin3  x  cos  #). 

23.  cos  4  #  =  4  cos4  #  4-  4  sin4  x  —  3. 

24.  sin  5  x  =  16  sin5  x  —  20  sin3  x  -f-  5  sin  a?. 

25.  cos  5  a?  =  16  cos5  x  —  20  cos3  x  +  5  cos  x. 

26.  sin  (a  +  ft  -f  y)  =  sin  a  cos  /?  cos  y 

+  cos  a  sin  /?  cos  y 
+  cos  a  cos  (3  sin  y 
—  sin  a  sin  /?  sin  y. 

27    tan(a-f-#+   x  _  tan  ot+ tan  ^  +  tan  y— tan  «tan  ft  tan  y 
1— tan  /3  tariy— tan  y  tan  a— tan  a  tan  ft' 


104  TRIGONOMETRY 

28.  2  sin  (a  -  ft)  cos  a  =  sin (2  a  -  £)  -  sin/3. 

29.  cos2(*:=1 


V2 
31. 


(*        )  V2 

32.  1 


4  1  -  tan  $ 


33.  tan  a  +  tan  ft  =   sinO  +  £) 

cos  a  cos  /3 

34.  cot«  +  cot^=s-H^±^l. 

sin  a  sin  /? 

35.  cos  (a  -f  ft)  cos  (a  —  ft)=  cos2 a  —  sin2)8. 


I.     COS0  = 


1  +  tan2  1  0 
37. 


cot  0  —  tan  ^ 

38.  tan  0  +  cot  0  =  2  cosec  2  0. 

39.  C0s"  +  sin'* 
cos  a  —  sin  a 

40. 


cos  (45°  -  0) 
41.   sin  (a  +  ft)  cos  («-/?) +cos  (a+ft)  sin  (a-)8)  =  sin  2 a. 

cos  a  -f-  cos  yS 


MISCELLANEOUS  EXERCISES  105 

43.  tan  2  «  —  tan  a  =  tan  a  sec  2  a. 

44.  sin«  =  2cos2K 

cot^a 

45.  cos  3  a  cos  a  +  sin  3  a  sin  a  =  cos  2  a. 


4, 

[sin  [£•-£]  4.  cos  [  ^-oTj    =  sin  a  +  cos  /?. 
\    A   )  \    *  yj 

47.   4  sin  0  sin  (60°  -  0)  sin  (60°  +  0)  =  sin  3  0. 
48. 


2sin0  +  sin20 

Solve  each  of  the  following  equations  for  all  values  of 
the  unknown  quantity  less  than  360°. 

49.  tan  0=  sin  0. 

50.  (3-4cos2a)cos2a=0. 

51.  sin  a  +  cos  a  cot  a  =  2. 

52.  tan  (45°  +  0)  =  3  tan  (45°  -0). 

53.  5  sin  0  =  tan  0. 

54.  1  -f  sin2  a  =  3  sin  a  cos  a. 

55.  2  cos  x  +  sec  x  =  3. 

56.  tan42/-4tan22/-h3  =  0. 

57.  sec  J3  4-  tan  ft  =  2. 

58.  2  sin  a  +  5  cos  a  =  2. 

59.  sin  5  x  -\-  sin  3  x  =  0. 

60.  cos  7  x  —  cos  a?  =  0. 

61.  tan60  =  l. 

62.  tan  20  tan  0  =  1. 

63.  sin40  +  sin20-f  cos0  =  0. 


106  TRIGONOMETRY 

64.  sin20-2cos0  +  2sin0-2  =  0. 

65.  sin20  +  sin30  =  2sin0cos20  — J. 

66.  tan20cot0-tan20  +  cot0-l  =  0. 

68.  sec2  0  esc2  0  +  4  =  4  esc2  0  +  sec2  0. 

69.  Given  tan  (3  =  u,  find  sin  2  /?. 

70.  Given  tan  0  =  esc  2  0,  find  cos  0. 

71.  Given  cos  x  =  -|,  find  cos  i  #. 

72.  Given  cos  x  =  f ,  find  tan  i  x  and  tan  2  a?. 

73.  Given  tan  2x  =  m,  find  tan  x. 

74.  If  «  +  p  +  y  =  180°,  show  that 

tan  a  +  tan  /?  +  tan  y  =  tan  a  tan  /?  tan  y. 

75.  If  a  +  ft  +  y  =  180°,  show  that 

cot  ^  +  cot  f±  +  cot  2  =  cot  -  cot  £  cot  2 . 

^  ^  2  222 

76.  If  a  +  £  +  y  =  180°,  show  that 

sin  a  +  sin  ft  +  sin  y  =  4  cos  £  cos  ^  cos  2« 

222 

Prove  the  following : 

77.  tan-'i  +  tan-1!^. 
^^S  4  54 

78.  tan"1  -  +  tan"1 

6 

79.  tan-1  k  +  tan-1 1  = 


1  —  kl 
80. 


81. 


MISCELLANEOUS   EXERCISES  107 

82.  sin-1 

83.  cos-1 


m  +  n  *  n 

m  _,    /    n 


n 
84.   cot-1    fe2~2  -  =  2csc-1fc. 


Solve  the  following  for  y: 
85. 


86.    sm-12/  +  sin-12;y  =  - 

2 


87. cot-1 3  y  • 

88.   tan'1  y  =  sin"1  a  +  cos"1  b. 

89.   Prove  that  in  a  plane  triangle,  right-angled  at  y, 


90.   In  a  right  triangle,  c  being  the  hypothenuse,  prove 
that 

.  91        c  —  b         91 

l  - 


91.  In  an  isosceles  triangle  in  which  a  =  6,  prove  that 

cos«  =  ^;    versy  =  ^2. 

92.  In  a  triangle  in  which  y  =  60°,  prove  that 


93.   Show  that  the  area  of  a  regular  polygon,  inscribed  in 

•    i       u  j*      •        .    Tir2     .     2?r 

a  circle  whose  radius  is  r,  is  -^-  sin  -- 


108  TRIGONOMETRY 

94.    Show  that  the  area  of   a  regular  polygon,  circum- 
scribed about  a  circle  whose  radius  is  r,  is  nr2  tan  -. 


95.  Show  that  the  area  of  a  regular  polygon  of  n  sides  is 
-j-  cot  -i    a  being  the  length  of  a  side. 

96.  Prove  that  the  areas  of  an  equilateral  triangle  and  of 
a  regular  hexagon,  of  equal  perimeters,  are  to  each  other  as 
2:3. 

97.  A  flagpole  125  ft.  high,  standing  on  a  horizontal 
plane,  casts  a  shadow  250  ft.  long  ;  find  the  altitude  of  the 
sun. 

98.  From  the  top  of  a  rock  that  rises  vertically  325.6  ft. 
out  of  the  water,  the  angle  of  depression  of  a  boat  was  found 
to  be  24°  35'  ;  find  the  distance  of  the  beat  from  the  rock. 

99.  A  balloon  is  directly  above  a  station  A.     From  a 
second  station  B,  in  the  same  horizontal  plane  with  A,  the 
angle  of  elevation  of  the  balloon  is  60°  30'.   If  AB  =  300  ft., 
find  the  height  of  the  balloon. 

—  *  100.  From  a  tower  64  ft.  high  the  angles  of  depression 
of  two  objects  in  the  same  horizontal  line  with  the  base  of 
the  tower  and  on  the  same  side  of  the  tower,  are  measured 
and  found  to  be  28°  14'  and  42°  47'  respectively;  find  the 
distance  between  the  objects. 

101.  Find  the  area  of  a  triangular  field  whose  sides  are 
48  rods,  62  rods,  and  74  rods. 

***  102.  The  earth  subtends  an  angle  of  17£"  at  the  sun  ;  find 
the  distance  of  the  sun  from  the  earth,  the  radius  of  the 
earth  being  3960  mi. 

103.  A  vertical  tower  makes  an  angle  of  113°  12'  with  the 
inclined  plane  on  which  it  stands  ;  and  at  a  distance  of  89  ft. 
8  in.  from  its  base,  measured  down  the  plane,  the  angle  sub- 
tended by  the  tower  is  23°  26'.  Find  the  height  of  the  tower. 


MISCELLANEOUS  EXERCISES  109 

104.  In  a  circle  of  radius  8,  find  the  area  of  a  sector  with 
an  arc  of  50°. 

105.  In  a  circle  of  radius  12,  find  the  area  of  a  segment 
with  an  arc  of  132°. 

106.  In  a  circle  of  radius  r,  find  the  area  of  a  sector  with 
an  arc  of  1  radian. 

107.  The  area  of  a  square  inscribed  in  a  circle  is  200  sq.  ft. ; 
find  the  area  of  an  equilateral  triangle  inscribed  in  the  same 
circle. 

'•*  108.    Find  the  area  of  a  triangle  of  which  two  sides  are 
6  +  V5  and  6  -  V5,  the  included  angle  being  32°  12'. 

109.  At  what  latitude  is  the  radius  of  the  circle  of  latitude 
equal  to  1  the  radius  of  the  earth  ? 

110.  From  the  top  of  a  lighthouse  85  feet  high,  standing 
on  a  rock,  the  angle  of  depression  of  a  ship  was  3°  38',  and 
at  the  bottom  of  the  lighthouse  the  angle  of  depression  was 
2°  43' ;  find  the  horizontal  distance  of  the  vessel  and  the 
height  of  the  rock. 

111.  At  a  point  directly  south  of  a  flagpole,  in  the  hori- 
zontal plane  of  its  base,  I  observed  its  elevation,  45°;  then 
going  east  200  ft.  its  elevation  was  35°.     Find  the  height 
of  the  flagpole. 

112.  A  castle  and  a  monument  stand  on  the  same  hori- 
zontal plane.     The  angles  of  depression  of  the  top  and  the 
bottom  of  the  monument  viewed  from  the  top  of  the  castle 
are  40°  32'  18"  and  80°  17'  46",  and  the  height  of  the  castle 
is  104f  feet.     Find  the  height  of  the  monument. 

113.  At  the  distance  a  from  the  foot  of  a  tower  the  angle 
of  elevation  a  of  the  top  of  the  tower  is  the  complement  of 
the  angle  of  elevation  of  a  flagstaff  on  top  of  the  tower; 
show  that  the  length  of  the  staff  is  2  a  cot  2  a. 


110  TRIGONOMETRY 

114.  A  flagstaff  a  feet  high  is  on  a  tower  3  a  feet  high ; 
the  observer's  eye  is  on  a  level  with  the  top  of  the  staff, 
and  the  staff  and  tower  subtend  equal  angles.     How  far  is 
the  observer  from  the  top  of.  the  staff? 

115.  Two  towers  on  a  horizontal  plane  are  120  feet  apart. 
A  person  standing  successively  at  their  bases  observes  that 
the  angular  elevation  of  one  is  double  that  of  the  other; 
but,  when  he  is  halfway  between  them,  the  elevations  are 
complementary.     Find  the  heights  of  the  towers. 

116.  An  observer  sailing  north  sees  two  lighthouses  8 
miles  apart,  in  a  line  due  west ;  after  an  hour's  sailing  one 
lighthouse  bears  S.  W.,  and  the  other  S.  S.W.     Find  the 
ship's  rate. 

117.  From  the  top  of  a  house  42  ft.  high,  the  angle  of 
elevation  of  the  top  of  a  pole  is  14°  26'  9";  at  the  bottom  of 
the  house  it  is  23°  21'  33" ;  find  the  height  of  the  pole. 

118.  From  the  top  of  a  hill  I  observe  that  the  angles  of 
depression  of  two  successive  milestones  in  the  horizontal 
plain  below,  in  a  straight  line  before  me,  are  14°  20'  24" 
and  5°  31'  14".     Find  the  height  of  the  hill. 

119.  Along  the  bank  of  a  river  is  measured  a  line  500 
ft.  in  length ;   the  angles  between  this  line  and  the  lines 
of  sight  from  its  extremities  to  an  object  on  the  opposite 
bank  are  53°  1'  7"  and  79°  44'  55".     Find  the  breadth  of 
the  river. 

120.  A  cape  bears  N.  by  E.  (1  of  90°  E.  of  K),  as  seen 
from  a  ship.     The  ship  sails  N.  W.  30  miles  and  then  the 
cape  bears   E.     How  far  is  it  from  the  second  point  of 
observation  ? 

121.  Find  the  height  of  a  precipice,  its  angles  of  eleva- 
tion at  two  stations  in  a  horizontal  line  with  its  base  being 
39°  30'  and  34°  15'  and  the  distance  between  the  stations 
being  145  ft. 


MISCELLANEOUS  EXERCISES  111 

122.  From  the  summit  of  a  hill,  360  ft.  above  a  plain, 
the  angles  of  depression  of  the  top  and  bottom  of  a  tower 
standing  on  the  plain  were  41°  and  54°  ;  required  the  height 
of  the  tower. 

123.  At  one  side  of  a  canal  is  a  flagstaff  21  feet  high 
fixed  on  the  top  of  a  wall  15  feet  high;  on  the  other  side 
of  the  canal,  at  a  point  on  the  ground  directly  opposite,  the 
flagstaff  and  the  wall  subtend  equal  angles.    Find  the  width 
of  the  canal. 

124.  From  the  top  C  of  a  cliff  600  feet  high,  the  angle  of 
elevation  of  a  balloon  B  was  observed  to  be  47°  22',  and  the 
angle  of  depression  of  its  shadow  S  upon  the  sea  was  61° 
10'  ;  find  the  height  of  the  balloon,  the  altitude  of  the  sun 
being  65°  31'  and  J5,  S,  C  being  in  the  same  vertical  plane 
and  the  sun  being  behind  the  observer. 

125.  At  each  extremity  of  a  base  AB  =  758  yards,  the 
angles  between  the  other  extremity  and  two  objects  C  and 
D  were  observed,  viz.  CAB  =  103°  50'  41",  DAB  =  53°  17' 
24",  DBA  =  S5°  47'  30",  and  CBA  =  ±6°  13'  27";  find  CD. 

126.  Show  that  if  r  be  the  radius  of  the  earth,  h  the 
height  of  the  observer  above  the  sea,  and  d  the  angle  of 

depression  of  the  horizon  ;  then  tan  d  =  —  ^—  - 


127.  A  privateer  lies  12.75  miles  S.  W.  of  a  harbor,  and 
a  merchantman  leaves  the  harbor  in  a  direction  E.  by  S.,  at 
the  rate  of  10  miles  an  hour  ;  on  what  course  and  at  what 
rate  must  the  privateer  sail  in  order  to  overtake  the  mer- 
chantman in  \\  hours  ? 

128.  From  the  top  of  a  hill  the  angles  of  depression  of 
two  objects  in  the  plain  at  its  base  were  observed  to  be  45° 
and  30°,  and  the  horizontal  angle  between  them  is  also  30°; 
find  the  height  of  the  hill  in  terms  of  the  distance  a  be- 
tween the  objects. 


112  TRIGONOMETRY 

V-^129.  The  topmast,  120  feet  above  the  water  line  of  a 
man-of-war  coming  into  port  at  the  rate  of  10  miles  an  hour, 
was  first  seen  on  the  horizon  at  8.45  A.M.  by  a  person  swim- 
ming near  the  water's  edge;  and  at  10.06  A.M.  she  cast 
anchor.  Find  an  approximate  value  for  the  radius  of  the 
earth. 

130.  A  railway  curve  which  is  a  circular  quadrant  has 
telegraph  poles  at  its  extremities  and  at  equal  distance 
along  the  arc,  the  whole  number  of  poles  being  10.  A  per- 
son in  one  of  the  extreme  radii  produced  and  at  a  distance 
of  300  feet  from  its  extremity,  sees  the  third  and  sixth 
poles  in  line.  Find  the  radius  of  the  curve. 


CHAPTER  IX 


DE  MOIVRE'S  THEOREM  WITH  APPLICATIONS 

103.  In   the   present   chapter   De   Moivre's   theorem  is 
introduced   with  some   of  its   applications,  including  the 
demonstration  of  the  fundamental  series  by  means  of  which 
we  may  calculate  the  trigonometric  tables. 

A  few  preliminary  considerations  pertaining  to  the  com- 
plex number  furnish  the  necessary  basis. 

104.  Geometric  representation  of  a  complex   number.     In 
algebra  it  is  shown  that  every  complex  number  can  be  re- 
duced to  the  form  a  +  b V—  1  or  a  +  ib  where  a  and  b  are 
any  real  numbers,  and  i=V  —  1. 

Having  given  a  complex  number  a  +  ib,  we  can  construct 
a  point  P  whose  coordinates  are  y 

a  and  b.  This  point  may  be 
considered  the  geometric  repre- 
sentation of  the  complex  num- 
ber. It  is  thus  seen  that  to 
every  complex  number  there  cor- 
responds a  point  in  the  plane. 

Conversely,  to  every  point  in 
the   plane    there   corresponds   a 
complex  number,  since  the  coor- 
dinates of  the  point  represent  the  two  elements,  a  and  6,  of 
the  complex  number. 

The  line  OP,  instead  of  the  point  P,  may  be  considered 
the  geometric  representation  of  the  complex  number. 

113 


X'- 


114  TRIGONOMETRY 

From  the  figure  it  is  evident  that 
a  =  r  cos  a, 

and  b  =  r  sin  a. 

Therefore  a  +  ib  =  r  cos  a  +  ir  sin  a. 

Hence  any  complex  number,  a  -f  ib,  can  be  reduced  to  the 

r(cos  a  +  t"  sin  a).  (1) 

The  form  (1)  includes  all  real  and  all  pure  imaginary 
numbers  as  special  cases.  Letting  a  =  0°  or  180°,  (1)  reduces 
to  r  or  —  r,  which  represents  any  real  number  since  r  repre- 
sents any  real  positive  number.  Letting  a  =  90°  or  270°, 
(1)  reduces  to  ±  ri,  any  pure  imaginary  number. 

The  angle  XOP  or  a  is  the  argument  of  the  complex  num- 
ber a  -f  ib.  It  is  determined  by  the  equation 

tana=*. 
a 

The  distance  OP  or  r  is  the  modulus  of  the  complex  num- 
ber a  +  ib.  It  is  determined  by  the  equation 

r—  Va2  -f-  62. 
105.    To  show  that 

[r  (cos  a  +  i  sin  «)]n  =  rn  (cos  n«  + 1  sin  wa), 

91  6eift<7  a  positive  vnieger. 

Squaring  the  quantity  r  (cos  a  -\-i  sin  a),  we  have 
[r  (cos  a  +  i  sin  a)]2  =  r 2  (cos2  «  —  sin2  a-\-2i  sin  a  cos  a) 

=  r2(cos  2  a  +  i  sin  2  a)  by  Art.  71.     (1) 

Multiplying  each  member  of  (1)  by  r  (cos  a  +  i  sin  a)  we 

have 

[r  (cos  a  + 1  sin  a)]3=  r3[(cos  2  a  cos  «  —  sin  2  a  sin  a) 
+  /(sin  2  a  cos  a  -f-  sin  a  cos  2  a)] 
=  j-3  (cos  3  a  +  i  sin  3  a)  by  Art.  68.  (2) 

From  (2)  we  have,  similarly, 

[r  (cos  a-\-i  sin  a)]4  =  r4  (cos  4  a  -f-  i  sin  4  a).          (3) 
Equations  (1),  (2),  and  (3)  have  the  form 

[r  (cos  «  +  i  sin  «)]tt  =  rn  (cos  na  +  i  sin  w«).  (4) 


DE  MOIVRE'S  THEOREM  WITH  APPLICATIONS   115 

Multiplying  both  members  of  (4)  by  r  (cos  a  +  i  sin  a),  we 
have 
[r  (cos  a+i  sin  a)]n+1  =  rn+1[cos  (w  +  1)  <*+i  sin  (n +!)«].    (5) 

Hence,  assuming  the  law  expressed  in  (4)  to  be  true, 
equation  (5)  shows  that  it  is  true  when  n  is  increased  by 
unity.  But  the  law  is  true  for  n  =  4,  by  equation  (3) ; 
hence  it  is  true  for  n  =  5.  Being  true  for  n  —  5,  it  must 
also  be  true  for  n  =±  6,  etc.  Hence  equation  (4)  is  true  for 
all  positive  integral  values  of  n. 

It  can  be  shown  that  equation  (4)  is  still  true  when  n  is  a 
negative  integer,  or  a  fraction. 

From  equation  (4)  it  is  seen  that  the  ?ith  power  of  a  com- 
plex number  is  a  complex  number  having  an  argument  n 
times  the  argument  of  the  given  number  and  a  modulus 
equal  to  the  nth  power  of  the  given  modulus. 

Letting  r=l,  equations  (1),  (2),  (3),  and  (4)  become 

(cos  a  +  i  sin  a)2  =  cos  2  a  +  *  sin  2  a  (5) 

(cos  a  +  i  sin  a)3  =  cos  3  a  +  i  sin  3  a  (6) 

(cos  a  +  i  sin  a)4  =  cos  4  a  +  i  sin  4  a  (7) 

(  cos  a  +  /  sin  a)n  =  cos  /ia  +  /  sin  /ia.  (8) 

The  last  equation  is  known  as  De  Moivre's  theorem. 

PROB.     Show  De  Moivre's  theorem  is  true  when  n  =  —  3. 

106.   Geometric    Interpretation, 

Since  each  of  the  complex  num- 
bers         cos  a  +  i  sin  a 

cos  2  a  +  i  sin  2  a 

cos  3  a  +  i  sin  3  a 

cos  4  a  +  i  sin  4  a 
has  a  modulus  equal  to  unity, 
the  lines  representing  these  num- 
bers terminate  in  points  lying 
on  the  circumference  of  a  circle  whose  radius  is  unity. 
The  arguments  of  any  two  consecutive  integral  powers  of 
cos  a  +  i  sin  a  differ  by  a,  hence  the  lines  representing  any 
two  consecutive  powers  differ  in  direction  by  a. 


116  TRIGONOMETRY 

107.  Applications  of  De  Moivre's  theorem.     De  Moivre's 
theorem  may  be  used  to  find  the  various  roots  of  unity,  to 
extract  any  root  of  a  complex  number,  to  obtain  the  sine 
and  cosine  of  any  multiple  of  an  angle,  and  to  expand  the 
sine  and  cosine  of  an  angle  in  a  series  of  powers  of  the 
angle. 

108.  To  find  the  cube  roots  of  unity. 

If  the  cube  roots  of  unity  are  real  numbers,  or  complex 
numbers,  we  may  assume,  by  Art.  104,  that 

VI  =  r  (cos  a  +  i  sin  a).  (1) 

Then,  cubing,  1  =  r3  (cos  3  a  4-  i  sin  3  a).  (2) 

Also  l  =  r3cos(3a-2n7r)  +  zV3sin(3a-2ri7r),   (3) 

n  being  an  integer. 

Equating  the  real  and  imaginary  parts  of  equation  (3),  we 
have 


and  ?*3  sin  (3  a  —  2  nir)  =  0. 

These  equations  of  condition  are  satisfied  when 
3  a  —  2  mr  =  0  and  r  =  1, 

whence  a  =2-^. 

3 

When  n  =  0,    1,     2,     3,     4,  etc. 

«  -  0,  £=,  if,  ^,  S-  etc.,  respectively.          (4) 

o        o        o    •  o 

Every  angle  in  this  series  is  coterminal  with  either  0,  -^, 

o 

4  7T 

or  —  -;  hence  all  the  values  of  sin  a  and  cos  a  are  obtained 
o 

by  using  only  the  first  three  values  of  a.  Substituting 
these  values  of  a  in  equation  (1),  and  remembering  that  r=l, 
we  have 


DE  MOIVRE'S  THEOREM  WITH  •  APPLICATIONS   117 


•^1=1  forn  =  0, 

•\/I  =  -i  +  »iV3         forn  =  l, 
</l  =  -  £  - 1  i  V3         f or  n  =  2. 

The  three  cube  roots  of  unity 
are  represented  geometrically  by 
P1?  P2,  and  P3. 

We  can  now  write  the  three 
cube  roots  of  any  real  number 
a,  for  letting  al5  «2,  and  a3  be  the  cube  roots  of  a,  we  have 

Oi  =</a- 1,  02  =%(-  i  +  H  V3),  a3  =^a  (-  i  -  t£  V3), 
where  -\/a  is  the  arithmetical  cube  root. 

PROBLEM.     Show  that  (—  1  -f  i  -J-  V3)3  =  1,  thus  justifying 
the  assumption  made  in  Eq.  1. 

109.    To  find  the  fifth  roots  of  unity. 

Let    -v/1  =  r  (cos  a  +  i  sin  a).  (1 ) 

Then,  raising  each  member  to  the  fifth  power, 

1  =  r5  (cos  5  a  +  i  sin  5  a).  (2) 

Also          1  =  r5  cos  (5  a  —  2  mi)  +  it6  sin  (5  a  —  2  nw),  (3) 

n  being  an  integer. 

Equating  the  real  and  imaginary  parts  of  equation  (3),  we 
have  r5  cos  (5  a  —  2  mi)  =  1, 

and  r6  sin  (5  a  —  2  mi)  =  0. 

These  equations  of  condition  are  satisfied  when 
5  a  —  2  mr  =  0,  and  r  =  1. 

Therefore  «  =  — . 

o 

When    w  =  0,    1,       2,       3,       4,     etc. 


-jr,    -jr,    -jr>  etc.,  respectively.  (4) 


118  TRIGONOMETRY 

Substituting  the  values   from   (4)  in  equation  (1),  and 
remembering  that  r  =  1,  we  have 

j->2  -v/1  =  1  for  n  =  0, 

P»,   ^ 


il  +  isini?  for  n  =  2, 
o  o 


5  5 

=  cos  — +  isin—  forn  =  4. 
5  5 

110.    To  extract  the  square  root  of  a  +  ib. 
Let     Va  +  ib  =  r  (cos  a  +  i  sin  a).  (1) 

Squaring,  a  +  ib  =  r2  (cos  2  a  +  i  sin  2  a).  (2) 

Also  a  +  ib  =  ^[003(2  a  —  2nir)+i  sin  (2  a  —  2  rnr)],  (3) 

where  n  is  any  integer. 

Equating  the  real  and  imaginary  parts, 

^008(2  a  -  2tt7r)=  a,  (4) 

r2sin(2a-2ri7r)  =  6.  (5) 

Squaring  and  adding,  we  have 

r4=a2  +  &2,  (6) 

and  r  =  Va2  +  &2.  (8) 

Equation  (8)  gives  the  value  of  r  in  terms  of  the  known 

numbers  a  and  6. 

From  equations  (4)  and  (7) 


whence  2  a  —  2  nir  —  cos' 

the  quadrant  of  2  a  —  2  n-ir  being  determined  by  (4)  and  (5). 


DE  MOIVRE'S  THEOREM  WITH  APPLICATIONS  119 

Then  a  =  ±cos-1        a  forn  =  0,  (9) 

Va2  +  b2 

and  «  =  TT  +    cos"1  —    a  for  n  =  1.  (10) 


The  values  of  a  for  n  =  2,  3,  4,  etc.,  are  coterminal  with 
the  values  of  a  in  equations  (9)  or  (10).  Hence  there  are 
only  two  values  for  sin  a  and  cos  a  ;  namely  those  for  which 
n  =  0  and  n  =  l. 

Substituting  these  values  of  a  in  equation  (1),  we  have, 
finally, 

Va  +  ib  =  \/a2  -f  62[~cos  /£  cos-1        CT 


and  Va-M&  =  -\/a2  +  62|cos  f  «•  +  4  cos'1        a 
L      V  Va2  + 

+  i  sin  f  TT  4  i  cos-1  -  -  -  \\  .        (12) 
V  Va2  +  6V  J 

These  values  of  Va  +  16,  while  more  complicated  than 
Va  H-  ib  itself,  are  nevertheless  in  the  standard  form  of 
complex  numbers, 

r  (cos  a  +  i  sin  a). 

111.    To  extract  the  Mh  root  ofa  +  ib. 

Let  -fya  +  ib  =  r  (cos  a  +  i  sin  a).  (1) 

Then        a  +  £6  =  r*  (cos  fca  +  i  sin  fca)  (2) 

=  r*[cos  (Tea  -  2  rnr)  +  i  sin(fca  -  2  WTT)],  (3) 

where  n  is  an  integer. 

Equating  the  real  and  imaginary  parts  of  equation  (3),  we 

have  r*  cos  (ka  —  2  HIT)  =  a,  (4) 

and  i*  sin  (&«  —  2  WTT)  =  6.  (5) 


120  TRIGONOMETRY 

-Squaring  and  adding  equations  (4)  and  (5),  we  have 

r2*  =  a2  +  b2.  (6) 

Then  r*  =  Va2  +  62,  (7) 

and  r=  ^a2  +  62.  (8) 

From  equations  (4)  and  (7),  we  have 

cos  (ka- 


or  fca  —  2  nir  =  cos'1 


the  quadrant  of  ka  —  2  mr  being  determined  by  the  signs  of 
(4)  and  (5). 

Therefore     «  =  ?^Z[  +  !  cos"1        a 
k        k 


whence  a  = 


«=^  +  icos-1 
k        k 


4?r      1 

c^-r+'-cos-1 


k        k  Va2  +  b* 


Substituting  the  values  for  r  and  a  as  found  in  equations 
(8),  (10),  (11),  (12),  etc.,  in  equation  (1),  we  have  the  several 
kih  roots  of  a  +  ib. 

When   A;  is  an  integer  there  are  k  roots.      Consecutive 

values   of    a,   as  given  by  (9),  differ  by  — ,  hence  all  the 

fc 

values  of  a  after  the  fcth  are  coterminal  with  one  of  the 
first  k  values. 

112.  To  express  sin  na  and  cos  na  in  terms  of  sin  a  and 
cosa. 

We  have 

cos  na  +  i  sin  na  =  (cos  a  +  i  sin  «)n. 


DE  MOIVRE'S  THEOREM  WITH  APPLICATIONS     121 

Expanding  the  second  member  by  the  binomial  theorem 
and  equating  the  real  and  imaginary  parts,  the  problem  is 
solved. 

Thus,  for  sin  4  a  and  cos  4  a  we  have 
cos  4  a  4-  i  sin  4  a  =  (cos  a  -f  i  sin  a)4  =  cos4  a, 
-h  4  i  cos3  a  sin  a  —  6  cos2  a  sin2  a  —  4  i  cos  a  sin3  a  +  sin4  a. 
Therefore     sin  4  a  =  4  cos3  a  sin  a  —  4  cos  a  sin3  a. 
and  cos  4  a  =  cos4  a  —  6  cos2  a  sin2  a  -f-  sin4  a. 

113.  Comparison  of  the  values  of  sin  a,  a,  and  tan  a,  a  being 
an  acute  angle.  Let  a  be  any  acute  angle  expressed  in  ra- 
dians. With  the  vertex  0  as 
a  center  and  any  radius  OB, 
describe  the  arc  EC.  Draw  AG 
and  BD  perpendicular  to  OB, 
and  join  B  with  O. 

The  area  of  the  triangle  OBC 
is    less    than    the   area  of    the 
sector  OBC,  and  the  sector  OBC  is  less  than  the  triangle 
OBD. 

But  since 

AC=  OC  sin  a,  BD  =  OB  tan  a,  and  arc  BC  =  OB  •  a,  Art.  17, 
the  area  of  the  triangle  OBC  is  equal  to  |  •  OB  •  OC  sin  a, 
the  area  of  the  sector  OBC  is  equal  to  ^  •  OB  •  OB  -  a, 
and  the  area  of  the  triangle  OBD  is  equal  to  |-  •  OB  -  OB  tan  «. 
Hence 


or  sin  a  <  a  <  tan  a. 


114.   Value  of for  small  values  of  a.     In  Art.  40  we 

a 

saw  that  sin  a  approaches  0  as  a  approaches  0.     The  value 

of  —  therefore  approaches  .  as  a  approaches  0. 
a  (J 


122  TRIGONOMETRY 

But  from  the  previous  article 

sin  a.  <  a  <  tan  a. 
Dividing  by  sin  a,  we  have 

a  1 

sin  a      cos  a 

sin  a 

or  1  > >  cos  a. 

a 

sin  a  , .      .  sin  a 

Since lies  between  1  and  cos  a,  must  approach 

a  a. 

1  as  a  approaches  0,  since  cos  a  approaches  1. 

Then  for  very  small  angles  sin  a  may  be  replaced  by  «, 

expressed  in  radians.    The  error  thus  introduced  is  so  small 

that  it  may  be  neglected  in  many  problems.     Thus,  to  five 

decimal  places, 

sin  1°  =  0.01745  1°  =  0.01745  radians 

sin  2°  =  0.03490  2°  =  0.03491  radians 

sin  3°  =  0.05234  3°  =  0.05236  radians 

sin  4°  =  0.06976  4°  =  0.06981  radians 

115.    To  develop  sin  a  and  cos  a  in  terms  of  a. 

By  De  Moivre's  theorem, 

cos  nO  +  i  sin  nO  =  (cos  0  -f  i  sin  0)n. 

On  expanding  the  second  member  by  the  binomial  theorem, 
we  have 
cos  nO  +  *  sin  n&  =  cosn  0  -{-in  cos""1  0  sin  6. 

^j2 — -cosn~2  0  sin20  —  ?!  — j^ '  cosn~30  sin30 

|n(n-l)(n~2)(n-8)cos,_4gsin4g  |  ...  (1) 

11 
Equating  the  imaginary  parts,  we  have  ' 

sin  n$  =  n  cos""1 0  sin  0 r^ cosw~3  0  sin3  0 

tt(n-l)(n-2)(n-3)(tt-4) 
+  _A & A A /cosn 

l£ 
Let  nO  =  a.     Then  equation  (2)  may  be  written 


DE  MOIVRE'S  THEOREM  WITH   APPLICATIONS     123 


sin  a  =  ?  cos"-1 0  sin  0  -  ^ — ^ ^  cosw~3  0  sin3 


0  |3 


---- 

"  " 


Let   a  remain  constant   while   n  increases   indefinitely. 
Then   0  necessarily  decreases  indefinitely,  since  n0  =  a,  a 

constant.     By    Art.    114,   when   0  approaches  0,  —  ^—  ap- 

proaches 1,  and  cos  6  approaches  1.     Making  these  substi- 
tutions in  equation  (4),  we  have 

a3      a5      a7 


Equating  the  real  parts  of  equation  (1),  we  have 
cos  n6  =  cos"  0  -  n^~1)  cos"-2  6  sin2  6 

e 

n(n-  l)(n-2)(tt-3)          **•-.•«*  /KN 

4.  _1  -  d\_^  —  /i  --  ?  Cosn~4  6  sm4  ^  —  ••-.  (5) 

By  the  same  process  as  above,  equation  (5)  becomes 


The  series  for  sin  a,  and  cos  a  are  convergent  for  all  finite 
values  of  a*  They  enable  us  to  compute  the  sine  and 
cosine  of  any  angle.  It  is  then  possible  to  construct  a  table 
of  natural  functions,  from  which  the  logarithmic  functions 
may  be  obtained.  In  using  these  series  a  must,  of  course, 
be  expressed  in  radian  measure. 

*  See  any  College  Algebra  on  the  convergency  of  series. 


124  TRIGONOMETRY 

116.  EXAMPLES 

1.  Find  the  four  fourth  roots  of  unity  by  De  Moivre's 
theorem. 

2.  Find   the  six  sixth  roots  of   unity  by  De   Moivre's 
theorem. 

3.  Find  the  square  root  of  5  —  3  i. 

SOLUTION.  — Let  V5  —  3  i  =  r  (cos  a  +  I'sin  a).  (1) 

Then  5  -  3 1  =  r2  (cos  2  a  +  i  sin  2  «) 

=  r2[cos  (2  a  -  2  WTT)  +isin(2a  — 2nir)]. 
Equating  the  real  and  the  imaginary  parts, 

r2  cos  (2  a  -  2  WTT)  =  5,  (2) 

and  r2  sin  (2  a  -  2  WTT)  =  -  3.  (3) 

Squaring  and  adding  (2)  and  (3),  we  have 

r*  =  34,  .  •.  r2  =  V&,  and  r  =  ^34.  (4) 

Then  cos  (2  a  —  2  nir)  = » 

V34 

and  2  a  -  2  nir  =  329°  2',  (5) 

the  quadrant  being  determined  by  (2)  and  (3). 

When  n  =  0,  a  =  164°  31'  ;  (6) 

»  =  !,«  =  844°  31'.  (7) 

Substituting  from  (4)  and  (6)  in  (1),  we  have 
\/5^3l  =  -  2.3271  +  .6446  L 
Substituting  from  (4)  and  (7)  in  (1),  we  have 
V5^3l  =  2.3271  -.6446i. 

4.  Find  the  square  root  of  3  -f  4  i. 

5.  Find  the  square  root  of  —  3  —  4  i. 

6.  Find  the  square  root  of  1  +  2 1. 

7.  Find  the  square  root  of  i. 

8.  Find  the  square  root  of  —  i. 

9.  Find  the  cube  root  of  2  —  3  i. 

SOLUTION.  —  Let  \/2  -  3  i  -  r  (cos  a  +  i  sin  a).  (1) 

Then  2  -  3  i  =rs  (cos  3  a  +  i  sin  3  a) 

(3a-2n7r)]. 


DE  MOIVRE'S  THEOREM  WITH   APPLICATIONS     125 

Equating  the  real  and  imaginary  parts, 

r3  cos  (3  a  -  2  WTT)  =  2  ;  (2) 

and  r8  sin  (3  a  -  2  nir)  =  -  3.  (3) 

Squaring  and  adding  (2)  and  (3),  we  have 

i*  =  13,  .-.  r8  =  Vl3~  and  r  =  \/13T  (4) 

2 
Then  cos  (3  a  -  2  WTT)  =-— , 

and  3  a  -  2  nir  =  303°  41' ;  (5) 

the  quadrant  being  determined  by  (2)  and  (3). 

When  n  =  0,  a  =  101°  14' ;  (6) 

n  =  l,  a  =  221°  14';  (7) 

n  =  2,  a  =  341°  14'.  (8) 

Substituting  from  (4)  and  (6)  in  (1),  we  have 
•v/2^3l  =  -  .2987  +  1.5041  i. 
Substituting  from  (4)  and  (7)  in  (1),  we  have 

v  2-3i  =  -  1.1530  -  1.0107  *. 
Substituting  from  (4)  and  (8)  in  (1),  we  have 

•2/2  -  3  i  =  1.4518  -  .4933  i. 
We  have  thus  found  the  three  cube  roots  of  2  —  3  i. 

10.  Find  the  cube  root  of  1  +  i. 

11.  Find  the  cube  root  of  —  1  +i. 

12.  Find  the  cube  root  of  2  -f  3  i. 

13.  Find  the  values  of  sin  3x  and  cos  3  a;  in  terms  of 
sin  a?  and  cos  a;. 

14.  Find  the  values  of   sin  5  a;  and  cos  5  a;  in  terms  of 
sin  x  and  cos  x. 

15.  Prove  by  De  Moivre's  theorem  that 

sin  a  =  2  sin  ~  cos  ~>  also  cos  a  =  cos2  ~  —  sin2  -• 
2,       L  &  & 

I      a        .    a\2 
SUGGESTION,    cos  a  +  i  sin  a  =  I  cos  -  +  i  sm  -  1 .  f 

16.  Show  that  cos  a  =  cos3  ^  —  3  cos  f  sin2 1, 

o  o  o 

sin  a  =  3  cos2  ^  sin  ^  —  sin3  £ 
o         o  o 


SPHERICAL    TRIGONOMETRY 

CHAPTER   X 
FUNDAMENTAL  FORMULAS 

117.  The  spherical  triangle.*  Spherical  trigonometry 
treats  of  the  relations  between  the  various  parts  of  a 
spherical  triangle  and  of  the  methods  of  solving  the  spheri- 
cal triangle. 

The  sides  of  a  spherical  triangle 
are  always  arcs  of  great  circles. 

Having  given  a  spherical  triangle 
ABC,  situated  upon  a  sphere  S,  a 
triedral  angle  0- ABC  may  be  formed 
by  passing  planes  through  0,  the 
center  of  the  sphere,  and  through 
the  sides  of  the  triangle. 

It  is  known  from  geometry  that  the  arc  AB  and  the 
angle  AOB  contain  the  same  number  of  degrees,  and  that 
the  angle  CAB  and  the  diedral  angle  C-AO-B  contain  the 
same  number  of  degrees. 

The  sides  and  angles  of  a  spherical  triangle  may  have 
any  values  between  0°  and  360°. 

A  triangle  having  one  or  more  of  its  parts  greater  than 
180°  is  called  a  general  spherical  triangle. 

A  triangle  having  each  of  its  parts  less  than  180°  is  called 
a  spherical  triangle. 

We  shall  consider  only  those  triangles  whose  parts  are 
each  less  than  180°. 

*  For  a  course  on  the  right  spherical  triangle  read  Arts.  117  and  126 
and  from  Art.  128  to  end  of  Chapter  XI. 

127 


128 


TRIGONOMETRY 


118.   Law  of  sines.     To  find  the  relation  between  two  sides 
of  a  spherical  triangle  and  the  angles  opposite. 


/3  >  90°,  6  >  90°. 


Given  a  spherical  triangle  and  its  accompanying  triedral 
angle ;  through  the  vertex  P  pass  planes  perpendicular  to 
OR  and  OQ,  intersecting  in  PF  a  line  perpendicular 
to  the  plane  ORQ. 

Then 


sin  ft 

w 

Also 

sin  a 

-§- 

*-»=& 

therefore 
Hence 
Likewise 

sin  a 

RP 

sin  b 
sin  a 

QP 

sin  a 

sin/3 

sin  6 

sn  y     sn  c 
Uniting  these  equations,  we  have 

sing  _  sin  b  _  sine 
sin  a     sin  p     sin  Y 


FUNDAMENTAL  FORMULAS 


129 


This  demonstration  applies  to  similar  figures  drawn  for 
all  possible  cases,*  hence  the  theorem  is  always  true. 


To  find  the  relation  between  the  three 

A 

J 
v 


119.  Law  of  cosines. 

sides  and  an  angle. 

Given  a  spherical  tri- 
angle and  its  accompany- 
ing triedral  angle ;  pass 
a  plane  through  the  ver- 
tex A  perpendicular  to 
OA,  intersecting  the 
planes  of  the  triedral 
angle  in  the  lines  AB, 
AC,  and  BC. 

b  <  90°,  c  <  90°  ;  a  <  180°,  a  <  180°. 

Then  AB  =  r  tan  c,  OB  =  r  sec  c,  AC  =  r  tan  b,  OC  =  r  sec  b. 
From  the  triangle  OBC,  by  Art.  90, 

B&  =  (r  sec  b)2  +  (r  sec  c)2  —  2(r  sec  &)  (r  sec  c)  cos  a.      (1) 
Likewise  from  the  triangle  ABC, 

BC2  =  (r  tan  b)2  +  (r  tan  c)2  -  2  (r  tan  6)  (r  tan  c)  cos  a.  (2) 
Subtracting  (2)  from  (1),  we  have 

0  =  r2  (sec2  6  -  tan2  6)  +  r2  (sec2  c  -  tan2  c) 

—  2  r2  sec  b  sec  c  cos  a  +  2  r2  tan  6  tan  c  cos  <*, 
which  reduces  to 

0  ==  1  —  sec  6  sec  c  cos  a  +  tan  b  tan  c  cos  «, 
or  cos  a  =  cos  b  cos  c  +  sin  b  sin  c  cos  a.  (3) 

Also  cos  b  =  cos  c  cos  a  -f-  sin  c  sin  a  cos  p,  (4) 

and  cos  c  =  cos  a  cos  6  +  sin  a  sin  6  cos  y.  (5) 

*  The  following  seven  cases  can  arise : 

(1)  3  sides  <  90°,   3  angles  <  90°         (4)  1  side  <  90°,  2  angles  <  90° 

(2)  3  sides  <  90°,   2  angles  <  90°         (5)  1  side  <  90°,  1  angle  <  90° 
(8)  2  sides  <  90°,   2  angles  <  90°         (6)  1  side  <  90°,  0  angle  <  90° 

(7)  0  side  <  90°,   0  angle  <90°. 

It  is  to  be  understood  that  all  parts  not  mentioned  are  greater 
than  90°. 


130 


TRIGONOMETRY 


120-    To  extend  the  law  of  cosines. 
In  the  derivation  of  the  formula 

cos  a  =  cos  b  cos  c  -f-  sin  b  sin  c  cos  a, 

b  and  c  were  less  than  90°,  while  a  and  a  were  less  than  180°. 
To  show  that  the  formula  is  true  in  general,  it  is  necessary 
to  consider  two  additional  cases : 

1st.   Both  b  and  c  greater  than  90°. 
2d.   Either  b  or  c  greater  than  90°. 

Since  ft  and  y  do  not  enter  the  formula,  they  may  have 
any  value  consistent  with  the  above  conditions. 

First.    Given    the    triangle    ABC    in 
which  b  >  90°  and  c  >  90°. 

Extend  the  sides  6  and  c  of  the  tri- 
angle ABC,  forming  the  lune  whose 
angle  is  a.  Then  in  the  triangle  A'BC, 
the  sides  A'B  and  A'C  are  each  less 
than  90°,  hence  by  Art.  119 
cos  a  =  cos  (180°  -  b)  cos  (180°  -  c) 

+  sin  (180°  -  6)  sin  (180°  -  c)  cos  «, 
or  cos  a  =  cos  b  cos  c  +  sin  b  sin  c  cos  a. 
Hence  the  law  of   cosines  holds  when 
both  b  and  c  are  greater  than  90°. 
Second.   Given  the  triangle  ABC} 
in  which  b  <  90°  and  c  >  90°. 

Extend  the  sides  a  and  c  of  the 
triangle  ABC,  forming  the  lune 
whose  angle  is  ft.  Then  in  the  tri- 
angle AB'C,  the  sides  AB'  and  AG 
are  each  less  than  90°,  and  the 
angle  B'AC  is  equal  to  180°  -  a. 

Then,  by  Art.  119, 
cos  (180°  -  a)  =  cos  b  cos  (180°  -  c) 

+  sin  b  sin  (180°  -  c)  cos  (180°  -  a), 
or  cos  a  =  cos  b  cos  c  -f  sin  b  sin  c  cos  a. 


FUNDAMENTAL   FORMULAS  131 

Hence  the  law  of  cosines  holds  when  either  b  or  c  is  greater 
than  90°.     The  law  of  cosines  is  therefore  true  in  general. 

121.  To  find  the  relation  between  one  side  and  the  three 
angles. 

Let  a,  b,  and  c  be  the  sides  of  any 
spherical  triangle,  and  a',  b'}  c'  the  sides 
of  its  polar  triangle. 

Applying  the  law  of  cosines  to  the 
polar  triangle,  we  have 
cos  a'  =  cos  b'  cos  c'  +  sin  b'  sin  c'  cos  a'. 

But 

a'  =  180°  -  a,     b'=  180°  -  /?,  etc. 

Therefore 
cos  (180°  -  a)  =  cos  (180°  -  /?)  cos  (180°  -  y) 

+  sin  (180°  -  £)  sin  (180°  -  y)  cos  (180°  -  a), 
or  cos  a  =  —  cos  /3  cos  y  +  sin  /?  sin  y  cos  a.          (1) 

Also  cos  (3  =  —  cos  y  cos  a  +  sin  y  sin  a  cos  6,          (2) 

and  cos  y  =  —  cos  a  cos  ft  +  sin  a  sin  /?  cos  c.  (3) 

122.  The   sine- cosine  law.     To  find  the  relation  between 
three  sides  and  two  angles. 

We  have 

cos  a  =  cos  6  cos  c  +  sin  b  sin  c  cos  a,  (1) 

and  cos  b  =  cos  c  cos  a  -f  sin  c  sin  a  cos  /?.  (2) 

Eliminating  cos  a  by  substitution, 

cos  b  =  cos  b  cos2  c  +  sin  b  sin  c  cos  c  cos  a  +  sin  c  sin  a  cos  /?. 
Transposing  and  factoring, 

cos  b  (1  —  cos2  c)  =  sin  b  sin  c  cos  c  cos  a  -f-  sin  a  sin  c  cos  /?. 
Eeplacing  (1  —  cos2  c)  by  sin2  c,  and  dividing  by  sin  c,  we 
have 

cos  b  sin  c  =  sin  b  cos  c  cos  a  +  sin  a  cos  ft. 

Kearranging  terms, 

sin  a  cos  /3  =  cos  6  sin  c  —  sin  6  cos  c  cos  «.  (3) 


132  TRIGONOMETRY 

Also        sin  6  cos  y  =  cos  c  sin  a  —  sin.  c  cos  a  cos  ft,  (4) 

and          sin  c  cos  a  =  cos  a  sin  b  —  sin  a  cos  6  cos  y.  (5) 

Interchanging  ft  and  y  and  consequently  b  and  c,  we  have 
from  (3) 

sin  a  cos  y  =  cos  c  sin  b  —  sin  c  cos  6  cos  a.  (6) 

Similarly  from  (4) 

sin  6  cos  a  =  cos  a  sin  c  —  sin  a  cos  c  cos  /?,  (7) 

and  from  (5) 

sin  c  cos  ft  =  cos  6  sin  a  —  sin  &  cos  a  cos  y.  (8) 

123.  To  find  the  relation  between  two  sides  and  the  three 
angles. 

Applying  the  sine-cosine  law  to  the  polar  of  the  given 
triangle,  we  have 

sin  a'  cos  ft'  =  cos  bf  sin  c'  —  sin  b!  cos  cr  cos  ex! . 
But  a'=  180°  -  a,  0'  =  180°-  6,  etc. 
Then 
sin  (180° -a)  cos  (180°-&)=cos  (180° -0)  sin  (180° -y) 

-sin  (180°- /?)  cos  (180°- y)  cos  (180°-  a). 

Therefore  sin  a  cos  6  =  cos  ft  sin  y  -f-  sin  ft  cos  y  cos  a.     (1) 

Also  sin  /?  cos  c  =  cos  y  sin  a  +  sin  y  cos  a  cos  &,     (2) 

sin  y  cos  a  =  cos  a  sin  /3  -f  sin  a  cos  /?  cos  c,     (3) 

sin  a  cos  c  =  cos  y  sin  ft  +  sin  y  cos  /?  cos  a,    (4) 

sin  ft  cos  a  =  cos  a  sin  y  4-  sin  a  cos  y  cos  b,     (5) 

sin  y  cos  b  =  cos  /3  sin  a  +  sin  /?  cos  a  cos  c.    (6) 

124.  To  find  the  relation  between  two  sides  and  two  anglest 
one  of  the  angles  being  included  between  the  given  sides. 

From  Art.  122 

sin  a  cos  ft  =  cosb  sin  c  —  sin  6  cos  c  cos  a. 
Dividing  this  equation  by 

sin  a  sin  ft  =  sin  b  sin  a,  Art.  118 


FUNDAMENTAL  FORMULAS  133 

member  by  member,  we  have 

,  Q  _  cot  b  sin  c  —  cos  c  cos  a 

sin  a 

Therefore  sin  a  cot  (3  =  cot  b  sin  c  —  cos  c  cos  a.    ,          (1) 

Similarly  sin  ft  cot  y  =  cot  c  sin  a  —  cos  a  cos  ft  (2) 

and  sin  y  cot  a  =  cot  a  sin  b  —  cos  6  cos  y.  (3) 

Interchanging  a  and   ft  and  consequently  a  and  6,  we 

have  from  (1) 

sin  /?  cot  a  =  cot  a  sin  c  —  cos  c  cos  ft  (4) 

Similarly  from  (2) 

sin  y  cot  ft  =  cot  b  sin  a  —  cos  a  cos  y,  (5) 

and  from  (3) 

sin  a  cot  y  =  cot  c  sin  6  —  cos  b  cos  a.  (6) 

125.   Formulas  independent  of  the  radius  of  the  sphere.     It 

will  be  noticed  that  r,  the  radius  of  the  sphere,  does  not 
enter  any  of  the  formulas  thus  far  developed ;  hence  they 
are  independent  of  the  radius  of  the  sphere,  and  may  be 
applied,  without  modification,  to  triangles  on  any  sphere. 
The  fact  is  serviceable  in  problems  of  Astronomy  and  Ge- 
odesy where  the  formulas  are  applied  to  triangles  situated 
upon  the  celestial  and  terrestrial  spheres. 


CHAPTER  XI 
SPHERICAL   RIGHT  TRIANGLE 

126.  The  spherical  right  triangle  is  a  spherical  triangle 
one  of  whose  angles  is  a  right  angle.     The  other  parts  may 
have  any  values  between  0°  and  180°. 

In  the  work  that  follows  the  angle  y  will  be  taken  as  the 
right  angle. 

127.  Formulas  for  the  solution   of  right  triangles.      The 
formulas  for  the  solution  of  any  spherical  right  triangle  are 
obtained  from  the  general  formulas  of  Chapter  X  by  letting 
y  =  90°.     We  thus  have 

from  eq.  (1)  Art.  118  sin  a  =  sin  c  sin  a 

from-eq.  (1)  Art.  118  sin  b  =  sin  c  sin  /3 

from  eq.  (5)  Art.  119  cos  c  =  cos  a  cos  b 

from  eq.  (1)  Art.  121  cos  a  =  sin  ft  cos  a 

from  eq.  (2)  Art.  121  cos  ft  =  sin  a  cos  b 

from  eq.  (3)  Art.  121  cos  c  =  cot  a  cot  ft 

from  eq.  (2)  Art.  124  cos  ft  =  tan  a  cot  c 

from  eq.  (3)  Art.  124  sin  b  =  cot  a  tan  a 

from  eq.  (5)  Art.  124  sin  a  =cot  ft  tan  b 

from  eq.  (6)  Art.  124  cos  a  =  tan  b  cot  c. 

128.  Direct  geometric  derivation  of  formulas.     Let  a  and  b 
be  the  sides  of  a  given  spherical  right  triangle,  a  and  ft  the 
angles  opposite,  and  c  its  hypotenuse. 

Let  0-ABC  be  its  accompanying  triedral  angle. 

Through  the  vertex  B  pass  a  plane  perpendicular  to  OA, 
intersecting  the  planes  of  the  triedral  angle  in  AB}  BC, 
and  CA. 

134 


SPHERICAL  RIGHT   TRIANGLE 


135 


Then  Z  BAO=a,  Z  BOC=a,  Z  AOC=b,  and  Z.AOB=c. 
Also  Z  jB(L4,  Z BOO,  ZCAO,^BAO  are  each  a  right  angle. 
From  the  triangle  ABC  we  have 
OB 

;sina?       (1) 


OB 
AC 

AC     OA     tan  6 

cos  a.  = =  -pr-  = , 

AB     AB     tan  c 

OA 
CB 

CB     OC     tana 
tana  =  __  =  ^=___. 

OC 

By  interchanging  a  and  /?  and  consequently  a  and  6,  or 
by  passing  a  plane  through  D  JL  to  OB  and  proceeding  as 
above,  we  have 


(2) 


(3) 


o 
cos/?  = 


sin  c 

tana 
-  --  , 
tan  c 


sin  a 
Also  from  the  figure. 

OA 

OA      OC     cos  b 
=  OB=0^=seca 
OC 

Dividing  (1)  by  (5)  and  reducing  by  (7)  we  have 


, 

COS  0 

and  by  interchange  of  letters  as  before 


(5) 
(6) 

(7) 
(8) 


sin  (3 


cos  a 
cos  a 


136 


TRIGONOMETRY 


Substituting  the  values  of  cos  a  and  cos  b  from  (8)  and  (9) 
in  (7),  we  have,  after  reduction, 

cos  c  =  cot  a  cot  ft.  (10) 

In  the  demonstration  of  formulas  (1)  to  (10)  the  parts 
a,  ft,  a,  b,  and  c  were  assumed  less  than  90°.  To  show  that 
these  formulas  are  true  in  general,  it  is  necessary  to  con- 
sider two  additional  cases :  viz.  (1)  when  one  side  and  the 
hypotenuse  are  each  greater  than  90°,  and  (2)  when  the  two 
sides  are  each  greater  than  90°. 

1.    Wlien  a  >  90°  and  c  >  90°. 

Let    ABC  be    the   given 
spherical      right      triangle. 
Draw  the  lune  BB'.     Then 
in  the  right  triangle  AB'C 
a  each  part  is  less  than   90°. 

and  formulas  (1)  to  (10)  are  applicable. 

sm(180°-a)  =  sin(180°-a) 
'      sin  (180°  -  c) 

sin  a 


From  (1) 


or 


sin  a  = 


sine 


which  shows  that  (1)  holds  when  a  and  c  are  each  greater 
than  90°. 

From  (2)       cos  (180°  -  a)  =         tan  b 


or 


cos  a  = 


tan (180° -c) 
tan  b 


tanc' 
which  shows  that  (2)  also  holds  in  this  case. 

Similarly  it  may  be  shown  that  formulas  (3)  to  (10)  hold 
when  a  and  c  are  each  greater  than  90°. 

2.    When  a  >  90°  and  b  >  90°. 

Let  ABC  be  the  given  spherical 
right  triangle.    Draw  the  lune  (7(7. 
Then  in  the  right  triangle  ABC'  c 
each  part  is  less  than  90°  and  for- 
mulas (1)  to  (10)  are  applicable. 


SPHERICAL  RIGHT  TRIANGLE 


137 


From  (1)   sin  (180° -  a)  =  Bin (180° -a) 


sine 


or 


sin  a  = 


sin  a 
sin  G 


which  shows  that  (1)  holds  when  a  and  b  are  each  greater 
than  90°. 

From  (2)  cos  (180°  -  «)  = 
tan  b 


tan  c 


or 


cos  a  = 


tan  c' 


which  shows  that  (2)  also  holds  in  this  case. 

Similarly  it  may  be  shown  that  formulas  (3)  to  (10)  hold 
when  a  and  6  are  each  greater  than  90°. 

129.  Sufficiency  of  formulas.     It  will  be  noticed  that  the 
ten  formulas  of  Arts.  127  and  128  contain  all  possible  com- 
binations of  the  five  parts  of  a  spherical  right  triangle,  taken 
three  at  a  time ;  hence  they  are  sufficient  to  solve  any  spheri- 
cal right  triangle  directly  from  two  given  parts. 

130.  Comparison  of  formulas  of  plane  and   spherical  right 
triangles.     By   rearranging  the   formulas   of  the   previous 
article,  the  analogy  between  the  formulas  of  the  plane  and 
the  spherical  right  triangles  is  made  apparent. 


IN  PLANE  RIGHT  TRIANGLES* 


a 
sin  a  =  — 


cos  a  =  - 
c 

tan«=- 
o 


sin  /?=* 

C 

cos/3  =  - 
c 

tan/?=- 
a 


sin  a  =  cos  ft    sin  ft  =  cos  a 
c2  =  a2  +  b2 


IN  SPHERICAL   RIGHT  TRIANGLES 


cos  a  = 


sin  a 
sin  c 
tan  6 
tanc 
tana 
sin  b 


sin  ft  = 
cos  ft  = 


sin  b 
sin  c 
tana 
tanc 
tan  b 
sin  a 


sn  «  = 


1  =  cot  a  cot  ft. 

*The   above  comparison  is  taken  from  Chauvenet's 
Spherical  Trigonometry." 


cos  b  cos  a 

cos  c  =  cos  a  cos  b 
cos  c  =  cot  a  cot  /?. 

Plane  and 


138 


TRIGONOMETRY 


131.  Napier's  Rules.  The  ten  formulas  used  in  the  solution 
of  spherical  right  triangles  can  all  be  expressed  by  means 
of  two  rules,  known  as  Napier's  rules 
of  circular  parts. 

Napier's  circular  parts  are  the 
sides  a  and  5,  the  complements  of  the 
angles  opposite  or  90°  —  a,  90°  —  /?, 
and  the  complement  of  the  hypot- 
enuse or  90°  —  c. 

They  are  usually  written 


co  a 


a,     b,     GO  a,     co/3,     coc. 

It  will  be  noticed  that  the  right  angle 
is  not  one  of  the  circular  parts. 

Let  the  five  circular  parts  be  placed 
in  the  sectors  of  a  circle  in  the  order 
in  which  they  occur  in  the  triangle. 
Whenever  any  three  parts  are  considered, 
it  is  always  possible  to  select  one  of 
them  in  such  a  manner  that  the  other  two  parts  will  either 
be  adjacent  to  this  part,  or  opposite  this  part.  The  part 
having  the  other  two  parts  adjacent  to  it  or  opposite  it  is 
called  the  middle  part. 

Thus  let  co  a,  6,  and  a  be  the  parts  under  consideration. 
Then  b  is  the  middle  part  and  co  a  and  a  are  adjacent  parts. 

If  co  c,  co  /?,  and  b  are  the  parts  under  consideration,  b  is 
the  middle  part  and  co  c  and  co  y8  are  opposite  parts. 

Napier's  rules  may  now  be  stated  as  follows : 

The  sine  of  the  middle  part  is  equal  to  the  product  of  the 
cosines  of  the  opposite  parts. 

The  sine  of  the  middle  part  is  equal  to  the  product  of  the 
tangents  of  the  adjacent  parts.* 

*  To  associate  cosine  with  opposite  and  tangent  with  adjacent,  it 
may  be  noticed  that  the  words  cosine  and  opposite  have  the  same 
vowels ;  likewise  the  words  tangent  and  adjacent. 


SPHERICAL  RIGHT  TRIANGLE  139 

132.  Theorem.    In  a   spherical  right   triangle,  a  side  and 
the  angle  opposite  terminate  in  the  same  quadrant. 

From  the  equation 

cos  a  =  cos  a  sin  ft 

it  is  seen  that  cos  a  and  cos  a  must  always  have  the  same 
sign,  since  sin  j3  is  always  positive.  Hence  a  and  a  termi- 
nate in  the  same  quadrant. 

133.  Theorem.     Of  the  three  parts  a,  b,  c,  if  any  two  ter- 
minate in  the  same  quadrant,  the  third  terminates  in  the  first 
quadrant  ;  if  any  two  terminate  in  different  quadrants,  the 
third  terminates  in  the  second  quadrant. 

This  follows  directly  from  the  equation 

cos  c  =  cos  a  cos  b 

by  noticing  that  if  any  two  of  the  quantities  cos  a,  cos  6, 
and  cos  c  have  like  signs,  the  third  is  positive  ;  if  any  two 
have  unlike  signs,  the  third  is  negative. 

134.  Two  parts  determine  a  triangle.     In  order  to  solve  a 
spherical  right  triangle  two  parts,  in  addition  to  the  right 
angle,  must  be  given.     Each  of  the  required  parts  should  be 
obtained  directly  from  the  given  parts. 

Thus,  given 

•6  =  50°,  c  =  110° 

we  have  cos  a  =  —  -  , 

cos  50°  ' 

cos  a  =  tan  50°  cot  110°, 


using  the  formulas  for  spherical  right  triangles,  or  Napier's 
Rules.  If,  in  the  solution  of  a  problem,  the  sine  of  any 
required  part  is  found  to  be  negative,  no  triangle  is  possible, 
since  no  part  of  a  spherical  triangle  can  be  greater  than  180°. 
Likewise  if  the  logarithmic  sine  or  cosine  of  any  required 
part  is  found  to  be  greater  than  zero,  the  triangle  is  impossi- 
ble, since  no  sine  or  cosine  is  numerically  greater  than  unity. 


140  TRIGONOMETRY 

135.  The  quadrant  of  any  required  part.  Since  the  parts 
of  a  spherical  triangle  may  have  any  value  between  0°  and 
180°,  it  is  always  necessary  to  determine  whether  the  required 
parts  are  greater  or  less  than  90°.  This  can  be  done  by  the 
theorems  of  Arts.  132  and  133. 

Thus,  given 

6  =  50°,    c=110°, 

we  have  a  >  90°,  a  >  90°,  and  ft  <  90°. 

The  quadrant  in  which  any  required  part  terminates  may 
also  be  determined  from  the  formula  used  in  calculating  that 
part,  by  observing  the  signs  of  the  functions  involved.  But 
when  the  unknown  part  is  determined  from  its  sine,  the  part 
terminates  in  both  quadrants,  giving  two  solutions,  unless 
limited  by  the  theorems  of  Arts.  132  and  133. 

Thus,  given 

6  =  50°,  c  =  110°, 

by  writing  the  signs  of  each  function  above  the  function,  we 
have 

cos  110° 

cos  a  =  —  -  , 

cos  50° 
cos  a  =  tan  50°  cot  110°, 


sin  110° 

Then  a  >  90°  and  a  >  90°,  since  their  cosines  are  negative, 
and  (3  <  90°  by  Art.  132. 

136.   Check  formula.     The  formula  containing  the  three 
computed  parts  may  always  be  used  as  a  check  formula. 

Thus,  having  given 

6  =  50°,  c  =  110°, 
the  check  formula  is 

cos  a  =  cos  a  sin    . 


SPHERICAL  RIGHT  TRIANGLE 


141 


137.   Solution  of  a  right  triangle. 

Given  b  =  77°  35'  16"  and  a  =  112°  19'  42" 


tan  a  =  ^=-^- 
cota 

6 
a 

log  sin  b 
log  cot  a 
log  tan  a 

a 

77°  35'  16" 
112°  19'  42'' 
9.98973  -  10 
9.61353  -  10 

0.37620 
67°  11'  30" 
112°  48'  30" 

cos  a 


+ 
tan  b 

log  cos  a 
log  tan  b 
log  cot  c 

c 

9.57969  -  10 
0.65740 

8.92229  -  10 
85°  13'  13'' 
94°  46'  47" 

cos  /3  =  cos  &  sin  a 


log  cos  6 

log  sin  a 

log  cos  j3 

§ 


9.33233  -  10 
9.96615  -  10 


9.29848  -  10 
78°  31'  53" 


Check 

+ 
cos  /3  =  tan  a  cot  c 


log  tan  a 
log  cot  c 
log  cos  /3 


0.37620 
8.92229  -  10 


9.29849  -  10 


138.  Two  solutions  or  the  ambiguous  case.  Whenever  a 
side  and  the  angle  opposite  are  given,  there  are  two  solutions. 

Thus  if  a  and  a  are  given,  the  only  formulas  by  which  6, 
/3,  and  c  can  be  determined  are 

sin  6  =  tan  a  cot  a, 


sin/?.; 


cos  a 
cos  a3 


sin  a 


Since  the  unknown  parts  are  obtained  from  their  sines, 
each  may  have  two  values,  giving  two  solutions,  the 
theorems  of  Arts.  132  and  133  not  restricting  the  values  of 
the  parts  to  one  solution. 

Having  found  the  two  values  for  each  part,  the  theorems 


142 


TRIGONOMETRY 


of  Arts.  132  and  133  determine  the  values  that  belong  to 
each  solution. 

Thus,  given        a  =  155°  27'  45"          to  find  b 
a  =  100°  21' 50"  c 

ft 


SOLUTION 


sin  b  =  tan  a  cot  a 


a 
a 

log  tan  a 
log  cot  a 
log  sin  b 
b 

155°  27'  45" 
100°  21'  50" 
9.65945  -  10 
9.26217  -  10 

8.92162  -  10 
4°  47'  20"  and 
175°  12'  40" 

stn 

cos  a 

P  —   _ 
cos  a 

log  cos  a 
log  cos  a 
log  sin  /3 
18 

9.25503  -  10 
9.95889  -  10 

9.29614  -  10 
11°  24'  22"  and 
168°  35'  38" 

sin  c  = 


sing 

-4- 

sina 


log  sin  a 
log  sin  a 
log  sin  c 
c 

9.61835 
9.99286 

-  10 
-10 

9.62549 
24°  58' 
155°   1' 

-  10 
18"  and 

42" 

Check 


sin  b  =  sin  c  sin  /3 


log  sin  c 
log  sin  /3 
log  sin  6 


9.62549  -  10 
9.29614  -  10 


8.92163  -  10 


FIRST  SOLUTION 

&!=  4°  47' 20" 
ft=  11°  24' 22" 
C  =  155°  1'42" 


SECOND  SOLUTION 

62  =  175°12'40" 
A  =  168°  35'  38" 
c2=  24°  58' 18" 


SPHERICAL  RIGHT  TRIANGLE 


143 


139. 


EXAMPLES 


Solve    the    following   spherical    right    triangles,   right- 
angled  at  y. 

9.   a  =  132°  25' 
a  =  107°  30' 


1.    6=    10°  32' 
a=    12°    3' 


2.  a  =   25°  18' 
b=   32°  41' 

3.  c  =  120°37' 
ft  =      9°  49' 

4.  c=   46°  40' 
a=   20°  50' 

5.  a  =  115°    6' 
b  =  123°  14' 

6.  a  =  112°  43' 30" 
c=   85°  10' 10" 

7.  a  =   15°  18' 20" 
c=   21°  30' 40" 

8.  6  =  168°  13' 45" 
c  =  150°   9' 20" 


140.  Quadrantal  triangles.     A   quadrantal   triangle   is   a 
triangle  one  of  whose  sides  is  90°.     Its  polar  triangle  is 
then  a  right  triangle.      The  solution   of  a  quadrantal  tri- 
angle is  effected  through  the  solution  of  its  polar  triangle. 

141.  Isosceles   triangles.     The  solution   of    an   isosceles 
triangle  is  effected  by  solving  the  two  equal  right  triangles 
formed  by  dropping  a  perpendicular  from  the  vertex  to  the 
base. 


10.  c=   80°   3' 20" 
0  =  135°  16'  30" 

11.  6=171°   3' 15" 
c=   12°  20' 30" 

12.  a=   35°  54' 20" 
a=   47°    6' 10" 

13.  b=   15°    2' 30" 
J3=   20°  11' 40" 

14.  «=   20°  26' 20" 

£  =   84°  41'  40" 

15.  a=   25°  41' 30" 
a=   34°  25' 40" 


144  TRIGONOMETRY 

• 

142.  EXAMPLES 

Solve  the  following  triangles, 

1.  a  =  117°  54' 30"  3.  £=153°  16' 

b=    95°  42' 20"  a=    19°  3' 

c=    90°  c=    90° 

2.  a  =   69° 45'  4.  a  =  159°33'40" 
(3=   94°  40'  6=    95°  18' 20" 
c=  90°  c=    90° 

5.  The  base  of  an  isosceles  triangle  is  51°  8'.     The  equal 
angles  are   each  41°  57'.      Find  the   equal   sides    and  the 
angle  at  the  vertex. 

6.  The  base   angles    of   an   isosceles   triangle   are  each 
100°  12' 30",   the   vertical   angle   is   50°  19' 40".     Find   the 
equal  sides  and  the  base. 


CHAPTER  XII 
OBLIQUE    SPHERICAL   TRIANGLE 

143.  In   the   present   chapter   the   general  formulas   of 
spherical   trigonometry,   already  developed  (Chap.    X)  are 
transformed  into  standard  formulas  adapted  to  logarithmic 
computation;  and  the  problem  of  the  solution  of  the  spheri- 
cal triangle  is  discussed. 

GENEEAL  SOLUTION 

144.  To  find  the  angles  when  the  three  sides  are  given. 
From  Art.  119,  we  have 

cos  a  =  cos  b  cos  c  +  sin  b  sin  c  cos  a. 

Therefore       cos  «  =  cos  a.-cos  b  cos  c.  (1) 

sin  b  sin  c 

But  2  sin2  i  a  =  1  —  cos  a 

_  sin  b  sin  c  —  cos  a  -f-  cos  b  cos  c 
sin  b  sin  c 

cos  a  —  cos  (6  —  c) 

sin  b  sin  c 
Applying  formula  (4)  of  Art.  73,  we  have 

sin  b  sin  c 
Letting  2s  =  a  +  b-\-c,  we  have 


8in  i  a= 


sin  '6  sin  c 
145 


146  TRIGONOMETRY 

Similarly  uniting  (1)  with 

2  cos2  1  a  =  1  4-  cos  a, 

we  have       2  cos2  4  «  =  sin  6  sin  c  +  cos  «  -cos  6  cos  c 

sin  b  sin  c 

cos  (b  4-  c)  —  cos  CT 
sin  6  sin  c 

2  sin.±a  +  b      c  sin-—  a 


sin  6  sin  c 


Therefore    cos^ a=Jsin *  «h  (*-«).  (3) 

*       sin  b  sin  c 

Uniting  (2)  and  (3), 


ten  1  q  =  J8in  0  ~  *)  Sin  ('  ~  C)  =        tan  **  (4) 

^  —  — 


sin  5  sin  (s  —  a)          sin  (s  —  a) 


Similarly  tan  *  p  =  Jsin  (J  ~  c>  f  <«  ~  ">=  .   ^n\.   (5) 

\         sin  .«  sin  fx  —  h\  sin  <^.c  —  A^       ^   ' 


and          tan  H  =  J8in  (^  ~  ")  8in  ( 
^  — 


sin  5  sin  (s  —  6)          sin  (s  —  b) 
tan 


sin  5  sin  (5  —  (?)  sin  (5  —  c) 

where          tan  r  =     /sin  (s  -  a)  sin  (s  -  fr)  sin  (3  -  c)  . 
v  sins 

145.    To  find  the  sides  when  the  three  angles  are  given. 
Following  the  method  of  the  last  article,  the  equation 
cos  a=  —cos  ft  cos  y  4-  sin  ft  sin  y  cos  a 


gives 


.  -    sin  6  sin  Y 


and       cos  J  a  =  J«»(*-P)  eo.  (J- 
^  sin  p  sin  -y 


and      tan  J  a  =  J  ~'°cts^Qcos  (^~a)    =  tan  /?  cos  (5  -a) 
^-  - 


ctQ 
cos(S-P) 


OBLIQUE   SPHERICAL  TRIANGLE  147 

where         2#  =  a 


tan  R 


146.   Delambre's  or  Gauss's  formulas  express  relations  be- 
tween the  six  parts. 
By  Art.  68 

sin  i  (a  +  (3)  =  sin  i  a  cos  i'/3  +  cos  ^  a  sin  i.  /?. 

Substituting  for  sin  \  a,  cos  1  a,  sin  1  y8,  and  cos  |  /8  their 
values  in  terms  of  the  sides  of  the  triangle,  and  simplifying, 
we  have 


sin  i  (a  1  ff)  =  sin  (s  ~  6)  +  sin  (*  ~  a^  A  sln  S  Sln  (.' 
sin  c  ^      sin  a  si 


2  sin    c  cos    c 


sin  b 

*r- 


Then          sin  \  (a  +  p)  =  cos  i  (a  -6)  .  CQs  ^  ^  (1) 

cos^c 

Similarly  sin  £  (a  -  p)  =  sin^g~^  .  cos  i  Y,  (2) 

sin^c 


and  cos  J-  (a  +  p)  =  .  8in  J  Y,  (3) 

COS      C 


and  co8i(a-p)  =  .sin|Y.  (4) 


147.   Napier's  analogies  express  relations  between  five  parts 
of  a  triangle.  They  are  easily  obtained  from  Gauss's  formulas. 
Dividing  (1)  by  (2),  Art.  146, 


8in|(a_p)     tan|(a-6) 
Dividing  (3)  by  (4), 

.          tan    c 


cos|(a-p) 


148  TRIGONOMETRY 

Dividing  (4)  by  (2), 

cotjH 


Dividing  (3)  by  (1), 


148.  Formulas  collected.  The  following  formulas  are  suf- 
ficient to  solve  a  spherical  triangle  when  any  three  parts 
are  given  : 

c  1nr 


sin  s  sin  (s  —  a)          sin  (s  —  a) 
—)—  =  tan 


sin  !(<*  +  /?)  _       tan^c 
sin  i  (a  —  ft)      tan  1  (a  —  6) 

cos  -«  +    )  tan  ±  c 


cos  i  (a  —  ./8)      tan  1  (a  +  6) 

sin  1  (a  +  6)  _       cot  ^-  y 
sin  i  (a—  6)      tani(a  —  ^) 


1-6)      tanf 

sin  a  _  sin  b 
sin  a      sin  /? 


VII 


Formula  I  is  used  to  determine  an  angle  when  the  three 
sides  are  given. 

Formula  II  is  used  to  determine  a  side  when  the  three 
angles  are  given. 

Formulas  III  and  IV  are  used  when  two  angles  and  the 

*  These  formulas  are  typical.  Other  formulas  of  the  same  type  are 
obtained  by  a  cyclic  change  of  letters. 


OBLIQUE   SPHERICAL  TRIANGLE  149 

included  side  are  given.  Formula  III  determines  ^  (a  —  b) 
and  formula  IV  determines  -j-  (a  -j-  b),  from  which  a  and  b 
are  obtained.  Either  formula  may  also  be  used  to  deter- 
mine the  side  c  when  the  other  two  sides  and  their  opposite 
angles  are  given. 

Formulas  V  and  VI  are  used  when  two  sides  and  the  in- 
cluded angle  are  given.  Formula  V  determines  ^  (a  —  ft) 
and  formula  VI  determines  J  (a  +  (3),  from  which  a  and  ft 
are  obtained.  Either  formula  may  also  be  used  to  deter- 
mine the  angle  y  when  the  other  two  angles  and  their  op- 
posite sides  are  given. 

Formula  VII  is  used  when  an  angle  and  the  side  opposite 
are  among  the  given  parts. 

149.  Whenever  the  formulas  I  to  VI  are  employed  in  the 
solution  of  a   spherical   triangle   as   indicated   above,   the 
quadrant  in  which  any  part  terminates  may  always  be  deter- 
mined by  noticing  the  signs  of  the  functions  involved. 

But  when  the  law  of  sines  is  employed,  two  values  are 
found  for  the  required  part.  This  leads  to  two  solutions 
unless  limited  to  one  solution  by  the  following  principles. 

150.  Theorem.     Half  the  sum  of  any  two  angles  is  in  the 
same  quadrant  as  half  the  sum  of  the  sides  opposite. 

This  follows  from  a  consideration  of  the  signs  of  the 
functions  involved  in 

cos  \(a  +  ft)  _       tan^-  c 
cos  %(a  —  ft)      tan  |(a  -f  b) ' 

Since  each  part  is  less  than  180°,  tan  1  c  and  cos  ^(«  —  ft) 
are  always  positive.  Hence  cos|(«  +  /?)  an(i  tan  i  (a +  6) 
must  always  have  the  same  sign.  Hence  i(a  +  ft)  and 
|-(a  +  b)  terminate  in  the  same  .quadrant. 

151.  Theorem.    A  side  which  differs  more  from  90°  than  an- 
other  side,  terminates  in  the  same  quadrant  as  its  opposite  angle. 


150 


TRIGONOMETRY 


We  have  from  Art.  119 

cos  a  —  cos  b  cos  c 


cos  a 


sin  b  sin  c 


If  a  differs  more  from  90°  than  6,  cos  a  is  numerically 
greater  than  cos  b.  Cos  a  is  also  greater  than  cos  b  cos  c, 
since  cos  c  is  not  greater  than  unity.  Hence  the  numerator 
of  this  fraction  has  the  same  sign  as  cos  a.  The  denominator 
being  always  positive,  cos  a  and  cos  a  have  the  same  sign. 
Hence  a  and  a  terminate  in  the  same  quadrant. 

The  negative  of  this  theorem  is  not  true. 

Thus  given,         a  =  165°,  b  =  120°,  ft  =  135°, 
a  terminates  in  the  second  quadrant,  since  a  differs  more 
from  90°  than  6. 

152.  Theorem.  An  angle  which  differs  more  from  90°  than 
another  angle,  terminates  in  the  same  quadrant  as  its  opposite 
side. 

This  follows  from 

sff  cosy 


sin  ft  sin  y 
by  considerations  similar  to  those  of  the  previous  article. 

Thus  given,      a  =  80°,  y  =  140°,  and  a  =  120°, 
c  terminates  in  the  second  quadrant,  since  y  differs  more 
from  90°  than  a. 


153.   Illustrative  examples. 
1.   Given  the  three  sides,  a  =  105°  27'  20",  6 
c  =  96°  53'  10",  to  find  a,  ft,  and  y. 


83°  14'  40", 


a 

105°  27'  20" 

b 

83°  14'  40" 

c 

96°  53'  10" 

2s 

284°  94'  70" 

s 

142°  47'  35" 

s—  a 

37°  20'  15" 

s-b 

59°  32  '55" 

s  —  c 

45°  54'  25" 

Check-:  s 

142°  47'  35" 

tan  r  =  Jsin  («-«)  sin  (s-&)sin  Q-c) 
*  sins 


tan  \  a  = 


tanr 


tan  \  8  = 


sin  (s  —  a) 

tanr 
sin  (s  —  b) 

tan  r 
sin  (s  —  c) 


OBLIQUE   SPHERICAL   TRIANGLE 


151 


log  sin  (s—  a) 

9.78284-10 

log  sin  (s  —  6) 

9.93553-10 

log  sin  (s—  c) 

9.85623-10 

colog  sin  s 

0.21846 

2  log  tan  r 

9.79306-10 

log  tan  r 

9.89653-10 

log  tan  A  a 

0.11369 

log  tan  A  /3 

9.96100-10 

log  tan  A  7 

0.04030 

i« 

52°  24'  55" 

i£ 

42°  25'  51" 

IT 

47°  39'  17" 

0E 

104°  49'  50" 

£ 

84°  51'  42" 

T 

95°  18'  34" 

Check 


^U2/      siiiK«-&) 

a  +  b 
a-b 
*(«'  +  ») 
*(*-&) 

K«-0) 

188°  42'    0 
22°  12'  40' 
94°  21'    0' 
11°   6'  20' 
9°  59'    4' 

log  sin  £  (a  +  &) 

log  tan  £(«-£) 
colog  sin  A  (a  —  6) 
log  cot  £  7 
log  tan  1  7 

9.99875  -  10 
9.24563  -  10 
0.71531 

9.95969  -  10 
0.04031 

2.   Given  two  sides  and  the  included  angle,  a  =  29°  18', 
b  =  37°  30',  y  =  51°  52',  to  find  a,  (3,  and  c. 


*»>«=!!!  J(,g+??*°K6-«) 


b 

37°  30' 

a 

29°  18' 

y 

51°  52' 

b-a 

8°  12' 

b  +  a 

66°  48' 

4  (ft  -  «) 

4°   6' 

i  (&  +  a) 

33°  24' 

17 

25°  56' 

log  sin  |(6  —  a) 

8.85429 

-10 

log  cot  A  7 

0.31310 

colog  sin  i(6  +  a) 

0.25926 

log  tan  £  (£  —  cc) 

9.42665 

-  10 

1  03  -  a) 

14°  57'  14" 

log  cos  A  (&  _  a) 
log  cot  \  7 
colog  cos  \  (6  +  a) 
log  tan  A  (|8  +  «) 
103  +  0) 

a. 

9.99889  -  10 
0.31310 
0.07839 

0.39038 
67°  51'    8" 
14°  57'  14" 

82°  48'  22" 
52°  53'  54" 

log  tan  \  (b  -  a) 
log  sin  A  (|8  +  a) 
colog  sin  A  (^3  —  «) 
log  tan  A  c 

c 

8.85540  -  10 
9.96671  -  10 
0.58831 

9.41042  -  10 
14°  25'  43" 
28°  51'  26" 

152 


TRIGONOMETRY 


Check 

sin  a  _  sin  6  _  sin  c 
sin  a     sin  j8     sin  7 

log  sin  a 
log  sin  a 

9.68965 
9.90177 

-10 
-10 

log  sin  6 
log  sin  /3 

9.78445-10 
0.99656  -  10 

log  sin  c 
log  sin  7 

9.68361 
9.89574 

-10 
-10 

9.78788 

-10 

9.78789  -  10 

9.78787 

-10 

3.  Given  two  sides,  and  an  angle  opposite  one  of  them, 
a  =  63°  24'  50",  b  =  17°  36'  40",  a  =  44°  48'  20",  to  find  ft  y, 
and  c. 


sin  a 


taDic  =  *'"}(g  +  ^tanK«-6) 


a 

63°  24'  50" 

log  sin  J  (a  -f 

&) 

9.71445  -  10 

a 

44°  48'  20" 

log  tan  K«  -  /3) 

9.57083  -  10 

b 

17°  36'  40" 

colog  sin  \  (a  — 

ft) 

0.62876 

a  +  b 

62°  25'   0" 

log  cot  \  7 

9.91404  -  10 

a-b 

27°  11'  40" 

i 

7 

50°  38'  0" 

\  (a  +  6) 

31°  12'  30" 

7 

101°  16'  0' 

£(a  _  5) 

13°  35'  50" 

log  sin  a 

9.95147  -  10 

log  sin  b 

9.48081  -  10 

log  sin  £  (a  +  £) 

9.83375  -  10 

colog  sin  a 

0.15200 

log  tan  £  (a  — 

&) 

9.38359  -  10 

log  sin  /3 

9.58428  -  10 

colog  sin  \  (a  —  /3) 

0.45735 

/3 

22°  34'  44"    . 

log  tan  , 

c 

9.67469  -  10 

a  —  /3 

40°  50'    6" 

• 

j  C 

25°  18'  20" 

«  +  /3 

85°  59'  34" 

c 

50°  36'  4(X 

!(*-/») 

20°  25'    3" 

i  (a  +  /3) 

42°  59'  47" 

Cftecfc 

* 

sin  ft  _  sin  c 

sin  /3     sin  7 

log  sin  6     9.48081  - 

10                       log  sin  c 

9.88810  -  10 

log  sin  £     9.58428- 

10                       log  sin  7 

9.99155  -  10 

9.89653  - 

10 

9.89655  -  10 

OBLIQUE   SPHERICAL  TRIANGLE 


153 


154.  Two  solutions.  There  are  two  solutions,  if  any,  when- 
ever two  sides  and  an  angle  opposite  one  of  them,  or  two  an- 
gles and  a  side  opposite  one  of  them,  are  given,  unless  limited 
to  one  solution  by  the  principles  of  Arts.  150, 151,  and  152. 

Thus,  having  given  £  =  45°  15' 12",  b  =  56°  49'  46",  a  = 
68°  52'  48",  to  find  a,  y,  and  c. 


sin  6 


a 

b 

68°  52'  48" 
56°  49'  46" 
45°  15'  12" 

a 

a  +  P 
a-p 

ci  +  b 
a-b 

fo-&) 

52°  19'  33" 
97°  34'  45" 
7°   4'  21" 
48°  47'  22" 
3°  32'  10" 
125°  42'  34" 
12°   3'   2" 
62°  51'  17" 
6°    1'31" 

127°  40'  27" 
172°  55'  39" 
82°  25'  15" 
86°  27'  50" 
41°  12'  38" 

log  sin  p 
log  sin  a 
colog  sin  b 
.    log  sin  a 
a 

9.85140 
9.96980 
0.07725 

9.89845 
52°  19'  33",  or 
127°  40'  27" 

log  sin  £(«  +  £) 
log  tan  $  (a  -  6) 
colog  sin  &  (a  —  /9) 
log  tan  £  c 

I« 

c 

9.87639 
9.02346 
1.20984 

9.99917 
9.02346 
0.18124 

0.10969 
52°  9'  35" 
104°  19'  10'' 

9.20387 
9°   5'    7" 
18°  10'  14" 

log  sin  £  (a  +  6) 
logtan£O-£) 
colog  sin  }  (a  —  ft) 
log  cot  1  7 
i7 
7 

9.94932 
8.79099 
0.97895 

9.94932 
9.94238 
0.97895 

9.71926 
62°21'1" 
124°  42'  2" 

0.87065 
7°  40'  16" 
15°  20'  32" 

154 


TRIGONOMETRY 


Check 


cosJ(a-/8)    _                 ; 

logcos|(a  +  j3) 
log  tan  %(a  +  6) 
cologcos£(ot  —  /3) 
log  tan  \  c 

9.81877 
0.29012 
0.00083 

8.79013 
0.29012 
0.12361 

0.10972 
•&)  tan^r 

9.20386 
a  4-  B) 

cos  J(a  -  b) 


FIRST  SOLUTION 

«  =    52°  19'  33" 

c  =  104°  19'  10" 

=  124°42'    2" 


SECOND  SOLUTION 

a  =  127°  40'  27" 
c  =  18°  10'  14" 
y  =  15°  20' 32" 


155.   Area  of  spherical  triangle.     Representing  the  area  of 
a  sphere,  S,  by  720  spherical  degrees,  it  is  demonstrated  in 
geometry  that  the  area  of  a  spherical  triangle,  A,  in  terms 
of  spherical  degrees,  is  equal  to  its  spherical  excess  ;  or 
>f=(a  +  p+y  —  180)  spherical  degrees 

-' 


log  cos  £(  a  +  6) 
log  tan  £(«  +  £) 
cologcos  £(o  —  &) 
log  cot  £  7 

9.65920 
0.05762 
0.00241 

9.65920 
0.20904 
0.00241 

9.71923 

0.87065 

Let  A'  and  S'  represent  the  area  of  the  triangle  and  the 
sphere  respectively,  in  terms  of  the  unit  in  which  r  is  ex- 
pressed. Since  the  ratio  of  the  area  of  the  spherical  tri- 
angle to  the  area  of  the  sphere  is  independent  of  the  units 
used,  we  have  A'  __A 


Therefore 

But  the  area  of  the  sphere   expressed  in  terms  of  r  is 
4  ?rr2,  therefore  the  area  of  the  triangle  is  given  by 

A'  = 


180 


156. 


EXAMPLES 


Solve  the  following   spherical  triangles   and   check  the 
results : 


OBLIQUE   SPHERICAL  TRIANGLE 


155 


1. 

6=   10°    O'lO" 
c  =  114°  40'  40" 
a=   92°  28'  20" 

11.   «=   61°    8' 

J3=   59°  12' 
y=   78°  25' 

2. 

b=   85°    4'  19" 
c  =  139°  58'  25" 
«=   12°  20'  31" 

12.    a=   76°  43'  15" 
b=   83°  35'  27" 
c=   98°  26'  38" 

3. 

c=   82°   3'   4" 
a=   70°  14'  12" 
0  =   84°  20'    9" 

13.  «  =  110°35' 
^8  =  135°  42' 

y  =  146°     8' 

4.  a=,  95°    3' 30" 
b  =  128°  38'  50" 
y  =  170°  52'  20" 

5.  o=   29°  18' 
6=   37°  30' 
y  =   51°  52' 

6.  a=   11°  21' 10" 
y  =   19°    0'20" 
6=   66°  19' 30" 

7.  y  =  179°  22' 11" 
a  =  148°  17' 17" 
6=  25°  39' 34" 

8.  a  =  108°    5' 18" 
6  =  170°  30' 46" 
c=   85°  50' 22" 

9.  a  =  105°  27' 20" 
6=   83°  14' 40" 
c=   96°  53' 10" 

10.   a=   34°  19' 30" 

6=   28°  37' 10" 
c=   22°  44' 40" 


14.  y=   11°  34' 10" 

6=   82°  56' 30" 
c=   27°    9' 40" 

15.  a=   32°   4' 10" 
/?  =  128°56'20" 
a=   39°  50' 30" 

16.  /?=   80°  40'   2" 
c=   75°  54'   0" 
6  =  100°  21'  28" 

17.  a=   21°    I'lO" 
y=   17°  22' 50" 
a=   14°  13 '30" 

18.  )8=   77°  44' 55" 
y=   92°  17 '24" 
a=   26°  29' 39" 

19.  a  =  114°  23'    9" 
0=   88°  41' 11" 
y=   79°    0'   4" 

20.  0=   99°   4' 12" 

y  =  106°   0'   9" 
a  =  161°    2' 10" 


156  TRIGONOMETRY 

21.  c  =  100°  10' 40"  23.    b=   42°  15' 20" 
a=   65°  20' 30"  c  =  127°   3' 30" 
y=   94°  30' 10"                         (3=   31°  44' 20" 

22.  a  =  103°  19'  50"  24.    y  =  127°   4' 10" 
/?=   92°  37' 30"  a=   88°  12'   0" 
y  =  128°  54' 20"  c  =  141°  20' 30" 

SOLUTION  WHEN  ONLY  ONE  PART  IS  REQUIRED 

157.  In  many  problems  of  astronomy  and  geodesy,  it  is  re- 
quired to  find  only  one  or  two  of  the  unknown  parts  of  a  spher- 
ical triangle,  the  other  unknown  parts  being  of  no  importance 
in  the  problem.     It  then  becomes  desirable  to  have  a  method 
whereby  the  required  parts  can  be  computed  without  being 
under  the  necessity  of  first  computing  any  part  not  desired. 

It  has  already  been  shown  that  any  angle  can  be  found 
directly  from  three  given  sides  (Art.  144),  and  that  any 
side  can  be  found  directly  from  three  given  angles  (Art.  145). 

It  is  evident  that  any  part  can  be  found  from  any  three 
given  parts  by  the  use  of  the  general  formulas  containing 
the  required  part  and  the  three  given  parts.  By  the  intro- 
duction of  auxiliary  quantities,  these  formulas  will  now  be 
adapted  to  logarithmic  computation. 

158.  Given  two  sides  and  the  included  angle,  to  find  any 
one  of  the  remaining  parts. 

Let  a,  b,  y  be  the  given  parts. 

First.    To  find  c. 

The  relation  between  a,  6,  y,  and  c  is  (Art.  119), 

cos  c  =  cos  a  cos  6  -f  sin  a  sin  b  cos  y.  (1) 

To  adapt  this  formula  to  logarithmic  computation,  let 

m  sin  M=  sin  b  cos  y,  (2) 

and  m  cos  M  =  cos  b.  (3) 


OBLIQUE   SPHERICAL  TRIANGLE 


157 


Then  eliminating  b  by  uniting  (1),  (2),  and  (3),  we  have 

cos  c  =  m  (cos  a  cos  M+  sin  a  sin  M ), 
or  cos  c  =  m  cos  (a  —  M ). 

From  (2)  and  (3) 

tan  Jf  =  tan  6  cos  y, 
and  from  (3)  and  (4) 

cos  b  cos  (a  —  M") 
cos  c  = »— = £t 


cos 


(4) 
(5) 
(6) 


Equations  (5)  and  (6)  enable  us  to  find  c. 
ILLUSTRATION.     Given  o=  75°  38'  20",  &  =  54°  54'  38",  and 
=  30°  17' 43". 


tan  M  =  tan  &  cos  7 


a 

75°  38'  20" 

6 

54°  54'  38" 

7 

30°  17'  43" 

log  tan  6 
log  cos  7 
log  tan  Jf 
M 

0.15333 
9.93623 
0.08956 

50°  51'  58" 

a-  M 

24°  46'  22" 

cos  c  =  cos  5  cos  (a -Jf) 
cos  Jf 


log  cos  b 
log  cos  (a  —  M) 
colog  cos  M 
log  cos  c 
c 

9.75956 
9.95808 
0.19987 

9.91751 
34°  12'  27" 

Second.    To  find  ft. 

The  relation  between  a,  b,  y,  and  ft  is  given  by  Art.  124, 
(equation  5),  from  which  we  have 

,  Q  _  cot  b  sin  a  —  cos  a  cos  y  x^ 

siny 

Multiplying  numerator  and  denominator  by  sin  b, 
,  ~_  cos  b  sin  a  —  sin  b  cos  a  cos  y 

sin  b  siiiy 
Again  using  equations  (2)  and  (3)  we  have 

m  (cos  M  sin  a  —  sin  Jf  cos  a) 
sin  6  sin  y 


(8) 


,  0 
cot  «  = 


or 


sin  b  sm  y 


(9) 
(10) 


158  TRIGONOMETRY 

As  before   tan  M  =  tan  6  cos  y.  (11) 

From  (2)  and  (10) 

.  (12) 


. 
sin  M 

Equations  (11)  and  (12)  enable  us  to  find  ft. 
Third.    To  find  y. 

By  interchanging  a  and  6  and  consequently  a  and  ft  in 
(11)  and  (12)  we  have,  calling  the  auxiliary  angle  N9 

tan  JV=  tan  a  cos  y  (13) 

and  cota  =  ™tysin(6-^)  (14) 

sm^" 

to  determine  a. 

159.  Two  parts  required.  It  will  be  noticed  that  the  same 
auxiliary  quantity  M  is  used  to  find  both  c  and  ft.  We  thus 
have  a  convenient  method,  much  used  in  astronomy,  for 
finding  a  side  and  an  angle  when  two  sides  and  the  included 
angle  are  given. 

For  finding  c  and  ft  we  have,  collecting  our  formulas, 

tan  M  =  tan  6  cos  y 


cos  M 

cot/?=cotysin(q-Jf)t 
sin  M 

Dividing  equation  (10)  of  Art.  158  by  equation  (4),  we  have 

cot/?_tan  (a  —  Jf) 
cos  c        sin  b  sin  y  ' 
which  serves  as  a  check  upon  c  and  ft. 
Similarly,  for  finding  c  and  a,  we  have 


OBLIQUE   SPHERICAL   TRIANGLE 

tan  N=  tan  a  cos  y 

cos  a  cos  (b  —  N} 
cos  c  = 5s L 

cos  N 


159 


sin  N 


160. 


PROBLEMS 


An  arc  of  lr  on  the  earth's  surface  is  equal  to  one  English 
geographical  mile. 

1.  Find  the  distance  between  Boston,  latitude  42°  21'  N., 
longitude  71°  41'  W.,  and  San  Francisco,  latitude  37°  48'  N. 
and  longitude  122°  28'  W.  P 


SOLUTION.  —  Let  APD  be  the  me- 
ridian of  Greenwich  from  which  longi- 
tude is  measured,  ABCD  the  equator, 
and  P  the  north  pole. 

Let  the  positions  of  San  Francisco 
and  Boston  be  represented  by  E  and  F 
respectively.  The  desired  distance  is 
EF. 


Then 


But 


Letting 


angle  DPE  =  122°  28', 

angle  DPF  =    71°  41', 

arc        EB=   37°  48', 

arc         CF=   42°  21', 

arc         PB  =  arc  PC  =  90°. 

PE=PB-EB  =  52°  12', 

PF  =  PC  -  FC  =  47°  39', 
Z  FPE  =  /.  DPE  -  Z  DPF  =  50°  47'. 
Z  FPE  =  7,  PE  =  a,  PF=  &, 


we  may  find  EF  or  c  by  the  formulas 

tan  M  =  tan  6  cos  7,  cos  c  =  cos&cos(« 

cos  M 


-M) 


160 


TRIGONOMETRY 


_  cos  b  cos  (a  —  M  ) 


cos  M 

a 

b 
y 

52°  12' 
47°  39' 
50°  47' 

log  cos  b 
log  cos  (a  —  M  ) 
colog  cos  M 
log  cos  c 
c 

9.82844  -  10 
9.97951  -  10 
0.08526 

log  tan  b 
log  cos  y 
log  tan  M 
M 
a-  M 

0.04023 
9.80089  -  10 

9.89321  -  10 
38°  33'  20",  or 
2313£  miles 

9.84112  -  10 
34°  44'  23" 
17°  27'  37'' 

2.  Find  the  distance  between  New  York  (lat.  40°  43'  K, 
long.  74°  0'  W.)  and  San  Francisco  (lat.  37°  48'  K,  long. 
122°  28'  W.). 

3.  Find  the  distance  between  Calcutta  (lat.  22°  33f  N., 
long.  88°  19'  E.)  and  Greenwich  (lat.  51°  29'  N.). 

4.  Find  the  distance  between  Baltimore  (lat.  39°  17'  N., 
long.  76°  37'  W.)  and  Calcutta. 

161.    Given  two  angles  and  the  included  side,  to  find  any 
one  of  the  rema  ining  parts. 
Let  a,  ft  c  be  the  given  parts. 

First.     To  find  y. 

The  relation  between  a,  ft  c,  and  y  is,  Art.  121,  eq.  (3), 

cos  y  =  —  cos  a  cos  ft  +  sin  a  sin  ft  cos  c.         -   (1) 
Let    ra  sin  M  =  cos  a,  (2) 

and       m  cos  M  =  sin  a  cos  c.  (3) 

Uniting  (1),  (2),  and  (3) 

cos  y  =  m  (—  sin  M  cos  /?  -f  cos  M  sin  /?), 
or  cos  y  =  m  sin  (ft  —  M).  (4) 

From  (2)  and  (3) 

cot  M  =  tan  a  cos  c,  (5) 


OBLIQUE   SPHERICAL  TRIANGLE  161 


and  from  (2)  and  (4) 

cos  a  sin  (8  — 


sinM 
Equations  (5)  and  (6)  enable  us  to  find  y. 


(6) 


Second.     To  find  a. 

The  relation  between  a}  (3,  c,  and  a  is  given  by  Art.  124, 
equation  (4),  from  which 

sin  8  cot  a  +  cos  c  cos  8  ^ 

cot  a  = •— .  ( 1 1 

sine 

,      _  sin  ft  cos  a  +  sin  a  cos  c  cos  ft  t  ,^ 

sin  a  sin  c 

Uniting  equations  (2),  (3),  and  (8), 

cot  a  =  — ^       " ""  — ^—      — ' ,  (9) 

sin  a  sin  c 

sin  a  sin  c 
From  (2),  (3),  and  (10) 

cot  M  =  tan  a  cos  c  (11) 

cot  c  cos  (8 — Jf)  /.,  0\ 

and  cota  = ^ L,  (12) 

cos  Jf 

from  which  a  is  found. 


To  find  b. 

From  (11)  and  (12)  by  interchanging  a  and  /?,  and  conse- 
quently a  and  6,  we  have,  calling  the  auxiliary  quantity  JV, 

cot  N=  tan  £  cos  c  (13) 


and  cot6  =  -  (14) 

cos  N 

to  determine  &. 

162.    Given  two  sides  and  an  angle  opposite  one  of  them,  to 
find  any  one  of  the  remaining  parts. 
Let  a,  bj  a  be  the  given  parts. 


162  TRIGONOMETRY 

First.     To  find  c. 

The  relation  between  a,  b}  a  and  c  is 

cos  a  =  cos  b  cos  c  +  sin  6  sin  c  cos  a.  (1) 

Let    m  sin  Jf  =  sin  b  cos  a,  (2) 

and        m  cos  Jf  =  cos  6.  (3) 

Then  cos  a  =  m  cos  (c  —  M ).  (4) 

From  equations  (2),  (3),  and  (4), 

tan  M  =  tan  b  cos  a  (5) 

and  cos  (c  -  M)  = Cos  a  cos  ^  (6) 

cos  b 

Equation  (5)  determines  M  and  equation  (6)  determines 
c  —  M.  Adding  these  values,  we  have  c. 

In  general  there  are  2  solutions  for  c.  We  may  limit  M 
to  positive  values  less  than  180°.  By  equation  (6)  c  —  M 
may  have  two  values,  numerically  equal  but  opposite  in 
sign,  giving  two  values  for  c  unless  the  sum  M  -\-  (c  —  M ) 
is  greater  than  180°  or  negative,  in  which  case  there  is  but 
one  solution. 

Second.   To  find  y. 

The  relation  between  a,  6,  a,  and  y  is 

sin  y  cot  a  =  cot  a  sin  b  —  cos  b  cos  y. 
Multiplying  by  sin  a  and  rearranging,  we  have 

sin  y  cos  a  +  cos  y  sin  «  cos  b  =  sin  a  cot  a  sin  6.          (7) 
Let  n  cos  JV=  cos  a,  (8) 

and  n  sin  ^=  sin  a  cos  6.  (9) 

From  (7),  (8),  and  (9)  we  have 

n  sin  (y  -f-  JV)  =  sin  a  cot  a  sin  6.  (10) 

Then  from  (8),  (9),  and  (10), 

tan  N==  tan  a  cos  6,  (11) 

and  sin  (y  -f  JV)  =  sin  N  cot  a  tan  6.  (12) 

Equation  (12)  determines  y  +  N  and  equation  (11)  deter- 
mines H.  Subtracting  the  second  value  from  the  first  gives  y. 


OBLIQUE   SPHERICAL   TRIANGLE  163 

In  general  there  are  two  solutions,  since  y  +  N  may  have 
two  values. 

Third.    To  find  ft. 

The  angle  ft  is  found  from 

.     0     sin  6  sin  a  ,*  o\ 

sm  ft  = : ,  (lo) 

sin  a 

which  in  general  gives  two  values. 

163.    Given  two  angles  and  a  side  opposite  one  of  them,  to 
find  any  one  of  the  remaining  parts. 
Let  a,  ft,  a  be  the  given  parts. 

First.    To  find  y. 

The  relation  between  a,  ft,  a,  and  y  is 

cos  a  =  —  cos  ft  cos  y  +  sin  ft  sin  y  cos  a.  (1) 

Let  m  sin  M  =  cos  ft,  (2) 

and  m  cos  M  =  sin  ft  cos  a.  (3) 

Then  cos  a  =  msin(y  —  M).  (4) 

From  (2),  (3),  and  (4) 

cot  M  =  tan  ft  cos  a,  (5) 

-,  .    /        •,  f^      cos  a  sin  J/"  //»>. 

and  sm(y  —  M)=—  —  (6) 

cosp 

Equations   (5)  and  (6)    determine   M  and  y  —  M,  from 
which  y  is  found. 

Two  solutions  may  be  possible,  as  in  Art.  162. 

Second.    To  find  c. 

The  relation  between  a,  ft,  a,  and  c  is 

sin  ft  cot  a  =  cot  a  sin  c  —  cos  c  cos  /?.  (7) 

Multiplying  by  sin  a  and  transposing, 

cos  a  sin  c  —  sin  a  cos  c  cos  /3  =  sin  /?  sin  a  cot  a.        (8) 

Let  n  cos  JV=  cos  a,  (9) 

and  w  sin  JV=  sin  a  cos  ft.  (10) 


164  TRIGONOMETRY 

Then  uniting  (8),  (9),  and  (10) 

n  sin  (c  —  JV)  =  sin  ft  sin  a  cot  a.  (11) 

From  (9),  (10),  and  (11) 

tan  N=  tan  a  cos  /?,  (12) 

and  sin  (c  —  N)  =  tan  ft  cot  a  sin  N.  (13) 

Equations  (12)  and  (13)  determine  N  and  c  —  JV,  from 
which  c  is  found. 

There  may  be  two  solutions. 

Third.  To  find  b. 

We  have  sin  b  =  sin  ?  sin  *•  (14) 

Bin« 

There  are  two  values  of  b  unless  restricted  to  one  solution 
by  the  principles  of  Arts.  150,  151,  and  152. 

164.  The  general  triangle.  The  parts  of  the  general  spheri- 
cal triangle  are  not  restricted  to  values  less  than  180°.  It 
can  be  shown  that  all  the  formulas  developed  for  the  oblique 
spherical  triangle  are  true  for  the  general  spherical  triangle 
if  the  double  sign  is  introduced  in  the  formulas  of  Arts. 
144, 145,  and  146. 


ANSWERS 

Art.  19  ;  Page  10 


5.  114°  35'  28",  286°  28'  40",  etc. 

6.  60°,  135°,  -  300°,  57°  17'  44",  36°  28'  31",  etc. 

7.  7ift. 

8.  2f  radians,  137°  .30'  34".          12.   247.16  E.  P.  M. 

25.882. 

9.  94°  3'  24".  13.   18.33  mi.  per  sec. 

10.  ifTrft.  14.   5236. 

11.  2.7216  radians.  15.   9.6  TT 

240  it  ft.  per  min. 

Art.  27  ;  Page  20 
1.   3d  and  4th.  2.   1st  and  4th.  3.   1st  and  3d 

7.  2d.  8.   3d.  20.    -?^.  21.   -^. 

o  o 

28.   sin  «r  =  -g%  V85,  cos  ^  =  —  ^  V85,  cot  a1  =  —  ^) 
sec  «!  =  —  ^  V85,  esc  06!  =  ^-  V85, 
sin  «2  =  —  -^-  V85,  cos  «2  =  -g7-  A/  85,  cot  02  =  —  -J-, 
sec  «2  =  -J-  V85,  esc  c^  =  —  1  V85. 

35.  if 

Art.  36  ;  Page  29 

1.  6  =  16.5  3.   c-=.869  5.   a  =  27° 
c  =  17.5                      a  =  .225  ft  =  63° 
ft  =  70°.                        a  =  15°.  b  =  72.6 

2.  a  =  1.44  4.   c  =  65  6.   a  =  18£° 
b  =  2.05                      a  =  23°  ft  =±  71|° 

0  =  55°.  0  =  67°  &  =  .  00867 

165 


166 


TRIGONOMETRY 


7.   a  =  346      8.   a  =  27°        9.   a  =  .029     10.   a 

c  =  507  ft  =  63°  b  =  .089  a  =  8.11 

0  =  47°  a  =  3.50  0=72°  6  =  9.48 


11.  6  =  5161 
c  =  5489 
£=70°  5' 

14.   c=. 00006294 
a  =  72°  26' 
=  17°  34' 


12.  a  =  .1384 
6  =  .2878 
ft  =  64°  19' 


13.  a  =  1.446 
c  =  1.719 
a  =  57°  17' 


15.   b  =  810.80  16.   17.  etc. 

a  =  47°  31'  32"  check  your 

=  42°  28'  28"  results. 


31.  Base  1331.1,  vertical  angle  149*  19'  10". 

32.  Base  angles  39°  23'  56",  base  1477.0. 

33.  Equal  sides  1622.9,  base  angles  37°  59' 37". 

34.  Equal  sides  219.75,  base  angles  68°  27'  19". 

35.  37.504. 

Art.  38;  Page  31 


1. 

290.83 

ft. 

7. 

2nrtai 

i  •• 

12. 

54.775 

mi. 

2. 

405.24 

ft. 

n 

13. 

153.72 

Ibs. 

3. 

263.92 

ft. 

8. 

130.99 

ft. 

38°  31' 

46". 

4. 

289.93 
21.442 

ft. 
ft. 

9. 
10. 

226.11 
2572.5 

ft. 
ft. 

14. 

739.38 
hi. 

mi. 

per 

5. 

2  nr  sii 

180° 

11. 

21.360 
22.638 

in. 
in. 

15. 

16  ft.  £ 

l& 

o  Z 

in. 

n 

6. 

132.52 

ft. 

69°  26' 

36". 

16. 

35°  16'. 

Art.  50;  Page  46 

2.  -  cos  10°,  -  sin  80°.  5.    cos  20°,  sin  70°. 

3.  -  cot  20°,  -  tan  70°.  6.    -  tan  80°,  -  cot  10°. 

4.  -.cot  20°,  -  tan  70°.  7.    -  sin  60°,  -  cos  30°. 
8.   cos  B.         9.    -tan  B.         10.    —  tan0.         11.    -cos 


ANSWERS 


16T 


11. 

12. 

15. 
18. 
20. 
21. 
35. 
37. 

33. 
39. 

41. 


Art.  56  ;  Page  51 

_,        tan  0  =  -  £  V5,       cot0  =  -£V5, 
sec  0  =  —  |  V5,        esc  0  =  f . 

sin  0  =  ^VUfl,       cos  0  =  T|^  V149,        tan  0  =  J^, 
sec  0  =  I V149,  esc  0  =  TV  VI49. 

z  =  30°.  16.   «  =  45°.  17.   z=0°,  60°. 

0  =  45°,  60°.  19.   sina  =  ±^-i.  +  £V5. 

Identity.  22.   0  =  120°.  24.   Identity. 

x  =  60°,  120°.  23.   0  =  30°.  25.  y  =  30°,  150. 

a  =  30°,  150°.  36.   a  =  30°,  60°,  120°,  150°. 

x  =  45°,  135°.  38.   x  =  45°. 

Art.  60;  Page  57 

tan2a-l.     34.   sinzTVl-sin2*.     35.    1  +  COSX 
tan2#  +  l  1— cos2  a; 

'         ±vrro2' 

_i_->/1   __  /«S  1  1 

-,  sec0 


a 


±  Vl  -  a2 


COS0 


cot0 


CSC0  = 


,  sec0 


COS0' 


43. 

50. 
55. 
58. 
59. 
60. 


?•      47.  0°,  90°.      48.   135°.     49.   0°,  30°,  60°,  150°. 

2 

30°,  150°.       51.  45°.       54.  36°  52'.        (See  tables.) 
46°  24',  90°.  57.   65°  54',  114°  6'. 

0°,  _  30°,  -  150°. 

tanu=±|V5,  ±iV2.  61.   195°,  345°. 

j,  ±|V2.  62.   54°,  234°. 


168 


TRIGONOMETRY 


Art.  70 ;  Page  66 

2.  ^V2(V3  -  1).     3.  iV2( V3  -  1).    4. 
3+V3 


1). 


5. 

1. 
2. 

3. 

32. 

34. 
35. 

1. 
3. 
5. 

7. 
10. 

13. 

14. 

20. 


3-V3 


7.  cos  ft.         8.   —  cot  a.         9.   —  cos  a. 


Art.  75 ;  Page  71 


*  V2  -  V2,  fV2  +  V2,  V3  -  2V2,  V3  +  2  V2. 
±  |V15,  -  |,  ±  |  V15,  ± 


20 


/O  O£        /?^O  O/W     H  fr 

.raj;;-  400  65!;2 

30°,  90°,  150°.  42.  0°,  30°,  90°,  150°. 

0°,  120°.  43.  0°,  90°,  120°. 


Art.  85 ;  Page  82 

60°,  120°,  420°,  etc.  2.  150°,  210°,  -  150°,  etc. 

45°,  225°,  -  135°,  etc.        4.  A. 


6. 


u 


±  Vl  4-  %2 
±  2  ?^  Vl  - 


±  Vw2  +  1 

9.  ±vr^: 
u.  i 


u 


15. 


±  2  Vw2  -  1 


—  v*±v^/l  —  u2.    21.  ± Vl— u2Vl— v2— uv. 


ANSWERS 


169 


-  V     2 

'  V   2u 

2Q        SoT  TTTTOV          T  — 

1            1 

30    z         ±a     - 

cot 
31.  1 

1     34.  «.: 

V 

cot"1  a     a 

0  rt2 

VI  +  a2 
32    VT=rf-aV 

.4  a  . 

\Xl-62_j_5Vl-a2 

35.  -i  (1  +  Va2—  1 
a& 

V62-l). 

Art.  101  ;  Page  98 

1.    6  =  24.5 
c  =  31.8 

y=78°. 

2.    a 

& 
c 

=  50° 
=  54.9 

=  58.6 

3.   a  =  61.4 
c  =  47.7 

y  =  48°. 

4.   /? 

y 

c 

=  48°  20' 
=  62°  40' 
=  76.1 

5.   0  =  68°  22' 
y  =  56°  58' 
a  =  107 

6.   « 

c 

=  40°  48' 
=  104°  52' 
=  141. 

7.   a  =  69°  22' 
0  =  38°  38' 

c  =  42.7 

8.   ft 

y 

c 

=  38°  37' 
=  31°  23' 
=  173. 

9.   a  =  44°  42' 
£  =  60°  20' 

y  =  74°  54' 

10.    a 

y 

=  72°  21'   • 

=  49°  38' 
=  58°  V 

11.   /3  =  34°  44'    or 
y  =  125°  16'  or 
c  =  35.8        or 

146°  16'      12.    a 
13°  44'                ft 
10.4                    6 

=  51°  44'  or  128°  16' 
=  84°  26'  or  7°  54' 
=  .431       or  .059 

170  TRIGONOMETRY 

Art.  101 ;  Page  98 

13.   y  =  39°  49' 50"  14.   a  =  13.081 

a  =  41.581  c  =  13.620 

c  =  41.432  /?  =  97°2'15" 

15.    a  =  90°  20'  34"  16.    a  =  126°  59'  18" 

ft  =  65°  17'  34"  ft  =  27°  29'  38" 

y=  24°  21' 50"  y  =  25°31'4" 

17.   a  =  1.9555  18.   y  =  77°  22'  16" 

a  =  43°  36'  35"  ft  =  51°  57'  20" 

ft  =  15°  37'  7"  a  =  83.732 

19.    a  =  87°  33' 58"  20.    a  =  78°  40'  32" 

y  =  18°  6'  24"  ft  =  41°  20'  47" 

b  =  1.4033  y  =  59°  58'  41'^ 

21.   a  =  32°  24'  0"  22.   Impossible. 

ft  =  55°  0'  28" 
y  =  92°  35'  32" 

23.  a  =  142°  28'  9"  or  21°  31'  11" 
ft  =  29°  31'  31"  or  150°  28'  29" 
a  =  .080746  or  .04862 

24.  y  =  76°  50'  20"  or  103°  9'  40" 
a  =69°  39'  35"  or  43°  20'  15" 
a  =  17.405  or  12.739 

25.    c  =  502.28  26.    a  =  96°  9'  32" 
b  =  300.25     .  ft  =  41°  11'  10" 

y  =  23°  7'  3"  y  =  42°  39'  18" 

27.  .4  =  150.  30.  ^  =  108.61 

33.  A  =  368.91.  35.   172.8  ft. 

36.   106.1ft.  37.   3.710  mi. 

38.      97.14 

124.59  39.   60.1ft. 
178.64 

40.   57.93  mi.  per  Lr.  -41.   3888.0ft. 

42.   6328.7  ft.  43.   239600  mi. 


ANSWERS 


171 


Art.  102  ;  Page  102 


1. 

4.   42°58/18//.     5.   35°  48' 35".     6.  13.754  in. 
7.  £.  8. 


2.  60',72M35'.    3.        ,         ,         , 


7T 

9* 

7T       7T       7T 

6'  3'  2 


55  7T       73  7T 


144     144 


9. 


4  7T 


9  '  3'    9 


12. 
50. 
51. 
54. 
56. 
57. 
59. 
60. 
61. 
62. 
63. 
64. 
66. 
67. 
68. 

69. 
73. 


11    tE    10?r    16£    22  TT    28  TT    34  TT    40  *• 

'    77 '    77  '    77  '    77  '    77  '    77  '    77  ' 

9fin.  49.   0°,  180°. 

30°,  150°,  210°,  330°,  45°,  135°,  225°,  315°. 

30°,  150°.      52.  tan-\(2  ±  V3).      53.   0°,  180°,  cos'1  f 

45°,  135°,  225°,  215°,  sin'1  Vf      55.   60°,  300°,  90°,  270°. 

60°,  120°,  240°,  300°,  45°,  135°,  225°,  315°. 

tan-1!.  58.   90°,  270°,  sin-1  (-|i). 

0°,  45°,  90°,  135°,  180°,  225a,  270°,  315°. 

0°,  45°,  60°,  90°,  120°,  180°,  225°,  240°,  270°,  300°,  315°. 

7i°,  37i°,  67$.°,  97i°,  127^°,  1571°,  etc. 

30°,  150°,  210°,  330°. 

90°,  270°,  70°,  110°,  190°,  230°,  310°,  350°. 

90°,  180°.  65.  210°,  330°. 

45°,  215°,  671°,  1571°,  247f,  337^°. 

60°,  120°,  240°,  300°,  18°,  54°,  90°,  126°,  162°,  etc. 

60°,  90°,  120°,  240°,  270°,  300°. 

70.  0,  ±V|.    71. 


172 


TRIGONOMETRY 


99.  532.  100.  50.0.  101.  1478.5. 

102.  93,470,000  mi.      103.  51.9  ft.         104.  27.925. 

105.  112.36.  106.  -•  107.  129.90  ft. 


106.  f 


108.  8.2596.  109.  70°  31'  43". 

111.  196.  112.  89.431. 

115.  90  ft.,  40  ft.    116.  13.66. 

118.  820.54.  119.  535.4. 

121.  567.3.  122.  132.6. 

124.  641.  125.  962.605. 

128.  a.  129.  4009. 


110.  5296  ft.,  251  ft. 
114.  «V2. 
117.  103.97. 
120.  25.43. 
123.  36.7  ft. 
127.  16.33  N.  75°  36'  E. 
130.  1674.3. 


Art.  116  ;  Page  124 


1.  1,  i,  —  1,  —  i. 
2.1,     i 


4.   ±(2  +  0- 

6.   ±  (1.272  +  .786  1). 

7. 

8.   _ 
10.  1.0842  +  .29051  i, 

-  .79370  +  .79370  1, 

-  .29051  -  1.0842  1. 


5.   ±(l-2t). 


Art.  139 ;  Page  143 


1.  /3  =  78°9'22" 
c  =  10°  45' 55" 
a  =  2°  14'  5" 

3.   6  =  8°  26' 14" 
a  =  120°  59' 19" 
a  =  95°  2' 10" 


2?    a=  41°  11'  53" 

£  =  56°  19' 56" 
c  =  40°  27' 11" 

4.  ^  =  75°  21' 53" 
6  =  44°  43' 49" 
a  =  14°  59' 33" 


ANSWERS  173 

5.    a  =  111°  23'  47'.'  6.   ft  =101°  38'  28" 
£  =  120°  40'  56"  a  =  112°  13'.  48" 

c  =  76°  33'  24"  b  =  102°  35'  26" 

7.   £  =  46°r28"  8.    a  =  68°  42' 11" 
b  =  15°  18'  0"  £  =  155°48'0" 

a  =  46°  2'  40"  a  =  27°  37' 26" 

9.  p  =  153°  31'  29"  or  26°  28'  31" 
c  =  50°  43'  22"  or  129°  16'  38" 
b  =  159°  48' 44"  or  20° 11' 16" 

Art.  142;  Page  144 


1.    a  =  117°  45'  28" 

2.    y  =  88°  23'  11" 

0  =  96°27'1" 

a  =  69°  48'  42" 

y  =  93°0'61" 

b  =  94°  22'  46" 

3.    a  =  8°  49'  46" 

4.    a  =160°  13'  48" 

y  =  28°3'4" 

0  =  105°  21'  16" 

b  =  10.6°  56'  53" 

y  =  104°  25'  45" 

5.   Sides,  32°  45'  6" 

6.   Sides,  112°  32'  20" 

angles,  105°  49'  32" 

base,  46°  15'  12" 

Art. 

156  ;  Page  154 

1.  0  =  11°0'47" 

2.   0  =  14°  53'  47" 

y=92°8'27" 

y  =  170°  26'  51" 

a  =  114°  42'  50" 

a  =  55°  56'  0" 

3.    «  =  71°r23" 

6.    a  =  24°  35'  10" 

y  =  84°  22'  25" 

c  =  43°  29'  48" 

6  =  82°1'30" 

0  =  154°  19'  20" 

8.   a  =  133°  28'  34" 

9.    a  =  104°  49'  50" 

0  =  169°  38'  12" 

0  =  84°  51'  42" 

y  =  132°  6'  14" 

y  =  95°  18'  24" 

10. 

a  =  84°  57'  8" 

0  =  57°  47'  44" 

v  =  43°4'36" 

174  TRIGONOMETRY 

14.    a  =  155°  14'  24"  or  21°  52'  40" 
£  =  25°  50'  58"  or  154°  9' 2" 
a  =  107°  34' 50"  or  58°  0' 44" 
15.   Two  solutions.  24.   No  solution. 

Art.  160 ;  Page  159 
2.  2229|  miles.  3.  4291J  miles. 


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